Principles of General Chemistry v. 1.0M
This is the book Principles of General Chemistry (v. 1.0M). This book is licensed under a Creative Commons by-nc-sa 3.0 (http://creativecommons.org/licenses/by-nc-sa/ 3.0/) license. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. This book was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz (http://lardbucket.org) in an effort to preserve the availability of this book. Normally, the author and publisher would be credited here. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally, per the publisher's request, their name has been removed in some passages. More information is available on this project's attribution page (http://2012books.lardbucket.org/attribution.html?utm_source=header). For more information on the source of this book, or why it is available for free, please see the project's home page (http://2012books.lardbucket.org/). You can browse or download additional books there.
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Table of Contents About the Authors................................................................................................................. 1 Acknowledgments................................................................................................................. 4 Dedication............................................................................................................................... 5 Preface..................................................................................................................................... 6 Chapter 1: Introduction to Chemistry .............................................................................. 9 Chemistry in the Modern World ................................................................................................................ 11 The Scientific Method.................................................................................................................................. 17 A Description of Matter ............................................................................................................................... 26 A Brief History of Chemistry....................................................................................................................... 46 The Atom....................................................................................................................................................... 58 Isotopes and Atomic Masses ....................................................................................................................... 69 Introduction to the Periodic Table............................................................................................................. 85 Essential Elements for Life .......................................................................................................................... 93 Essential Skills 1 ........................................................................................................................................... 98 End-of-Chapter Material ........................................................................................................................... 118
Chapter 2: Molecules, Ions, and Chemical Formulas................................................. 123 Chemical Compounds ................................................................................................................................ 125 Chemical Formulas .................................................................................................................................... 150 Naming Ionic Compounds ......................................................................................................................... 166 Naming Covalent Compounds .................................................................................................................. 183 Acids and Bases .......................................................................................................................................... 211 Industrially Important Chemicals ............................................................................................................ 223 End-of-Chapter Material ........................................................................................................................... 235
Chapter 3: Chemical Reactions....................................................................................... 238 The Mole and Molar Masses...................................................................................................................... 240 Determining Empirical and Molecular Formulas ................................................................................... 261 Chemical Equations.................................................................................................................................... 291 Mass Relationships in Chemical Equations ............................................................................................. 309 Classifying Chemical Reactions ................................................................................................................ 338 Chemical Reactions in the Atmosphere................................................................................................... 368 Essential Skills 2 ......................................................................................................................................... 381 End-of-Chapter Material ........................................................................................................................... 390
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Chapter 4: Reactions in Aqueous Solution................................................................... 402 Aqueous Solutions...................................................................................................................................... 404 Solution Concentrations............................................................................................................................ 419 Stoichiometry of Reactions in Solution ................................................................................................... 439 Ionic Equations........................................................................................................................................... 453 Precipitation Reactions ............................................................................................................................. 459 Acid–Base Reactions .................................................................................................................................. 473 The Chemistry of Acid Rain ...................................................................................................................... 502 Oxidation–Reduction Reactions in Solution ........................................................................................... 509 Quantitative Analysis Using Titrations.................................................................................................... 531 Essential Skills 3 ......................................................................................................................................... 544 End-of-Chapter Material ........................................................................................................................... 552
Chapter 5: Energy Changes in Chemical Reactions.................................................... 559 Energy and Work........................................................................................................................................ 561 Enthalpy ...................................................................................................................................................... 576 Calorimetry................................................................................................................................................. 621 Thermochemistry and Nutrition.............................................................................................................. 641 Energy Sources and the Environment ..................................................................................................... 654 Essential Skills 4 ......................................................................................................................................... 671 End-of-Chapter Material ........................................................................................................................... 680
Chapter 6: The Structure of Atoms ............................................................................... 685 Waves and Electromagnetic Radiation .................................................................................................... 687 The Quantization of Energy ...................................................................................................................... 697 Atomic Spectra and Models of the Atom ................................................................................................. 709 The Relationship between Energy and Mass........................................................................................... 732 Atomic Orbitals and Their Energies ......................................................................................................... 746 Building Up the Periodic Table................................................................................................................. 773 End-of-Chapter Material ........................................................................................................................... 796
Chapter 7: The Periodic Table and Periodic Trends.................................................. 802 The History of the Periodic Table............................................................................................................. 804 Sizes of Atoms and Ions ............................................................................................................................. 815 Energetics of Ion Formation ..................................................................................................................... 833 The Chemical Families............................................................................................................................... 867 Trace Elements in Biological Systems...................................................................................................... 893 End-of-Chapter Material ........................................................................................................................... 901
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Chapter 8: Ionic versus Covalent Bonding................................................................... 904 An Overview of Chemical Bonding........................................................................................................... 906 Ionic Bonding ............................................................................................................................................. 908 Lattice Energies in Ionic Solids................................................................................................................. 916 Lewis Electron Dot Symbols ...................................................................................................................... 937 Lewis Structures and Covalent Bonding.................................................................................................. 942 Exceptions to the Octet Rule..................................................................................................................... 978 Lewis Acids and Bases................................................................................................................................ 988 Properties of Covalent Bonds ................................................................................................................... 995 Polar Covalent Bonds............................................................................................................................... 1009 End-of-Chapter Material ......................................................................................................................... 1021
Chapter 9: Molecular Geometry and Covalent Bonding Models ........................... 1025 Predicting the Geometry of Molecules and Polyatomic Ions .............................................................. 1027 Localized Bonding and Hybrid Atomic Orbitals.................................................................................... 1069 Delocalized Bonding and Molecular Orbitals ........................................................................................ 1090 Polyatomic Systems with Multiple Bonds ............................................................................................. 1134 End-of-Chapter Material ......................................................................................................................... 1149
Chapter 10: Gases ............................................................................................................ 1154 Gaseous Elements and Compounds ........................................................................................................ 1156 Gas Pressure.............................................................................................................................................. 1165 Relationships among Pressure, Temperature, Volume, and Amount ................................................ 1181 The Ideal Gas Law..................................................................................................................................... 1191 Mixtures of Gases ..................................................................................................................................... 1219 Gas Volumes and Stoichiometry ............................................................................................................ 1230 The Kinetic Molecular Theory of Gases ................................................................................................. 1241 The Behavior of Real Gases ..................................................................................................................... 1265 Essential Skills 5 ....................................................................................................................................... 1278 End-of-Chapter Material ......................................................................................................................... 1284
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Chapter 11: Liquids......................................................................................................... 1290 The Kinetic Molecular Description of Liquids....................................................................................... 1292 Intermolecular Forces ............................................................................................................................. 1297 Unique Properties of Liquids .................................................................................................................. 1320 Vapor Pressure ......................................................................................................................................... 1334 Changes of State ....................................................................................................................................... 1351 Critical Temperature and Pressure ........................................................................................................ 1370 Phase Diagrams ........................................................................................................................................ 1377 Liquid Crystals.......................................................................................................................................... 1386 Essential Skills 6 ....................................................................................................................................... 1395 End-of-Chapter Material ......................................................................................................................... 1400
Chapter 12: Solids ........................................................................................................... 1403 Crystalline and Amorphous Solids ......................................................................................................... 1405 The Arrangement of Atoms in Crystalline Solids ................................................................................. 1412 Structures of Simple Binary Compounds .............................................................................................. 1426 Defects in Crystals.................................................................................................................................... 1442 Correlation between Bonding and the Properties of Solids ................................................................ 1463 Bonding in Metals and Semiconductors ................................................................................................ 1476 Superconductors ...................................................................................................................................... 1494 Polymeric Solids....................................................................................................................................... 1503 Contemporary Materials ......................................................................................................................... 1511 End-of-Chapter Material ......................................................................................................................... 1521
Chapter 13: Solutions ..................................................................................................... 1525 Factors Affecting Solution Formation ................................................................................................... 1527 Solubility and Molecular Structure........................................................................................................ 1539 Units of Concentration ............................................................................................................................ 1559 Effects of Temperature and Pressure on Solubility.............................................................................. 1577 Colligative Properties of Solutions......................................................................................................... 1590 Aggregate Particles in Aqueous Solution .............................................................................................. 1630 End-of-Chapter Material ......................................................................................................................... 1639
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Chapter 14: Chemical Kinetics ..................................................................................... 1643 Factors That Affect Reaction Rates ........................................................................................................ 1646 Reaction Rates and Rate Laws................................................................................................................. 1656 Methods of Determining Reaction Order .............................................................................................. 1679 Using Graphs to Determine Rate Laws, Rate Constants, and Reaction Orders .................................. 1713 Half-Lives and Radioactive Decay Kinetics............................................................................................ 1721 Reaction Rates—A Microscopic View ..................................................................................................... 1735 The Collision Model of Chemical Kinetics ............................................................................................. 1749 Catalysis .................................................................................................................................................... 1765 End-of-Chapter Material ......................................................................................................................... 1776
Chapter 15: Chemical Equilibrium .............................................................................. 1785 The Concept of Chemical Equilibrium ................................................................................................... 1787 The Equilibrium Constant ....................................................................................................................... 1794 Solving Equilibrium Problems ................................................................................................................ 1826 Nonequilibrium Conditions .................................................................................................................... 1852 Factors That Affect Equilibrium ............................................................................................................. 1869 Controlling the Products of Reactions................................................................................................... 1891 Essential Skills .......................................................................................................................................... 1900 End-of-Chapter Material ......................................................................................................................... 1903
Chapter 16: Aqueous Acid–Base Equilibriums .......................................................... 1911 The Autoionization of Water .................................................................................................................. 1913 A Qualitative Description of Acid–Base Equilibriums .......................................................................... 1926 Molecular Structure and Acid–Base Strength....................................................................................... 1958 Quantitative Aspects of Acid–Base Equilibriums .................................................................................. 1970 Acid–Base Titrations................................................................................................................................ 1999 Buffers ....................................................................................................................................................... 2030 End-of-Chapter Material ......................................................................................................................... 2058
Chapter 17: Solubility and Complexation Equilibriums ......................................... 2062 Determining the Solubility of Ionic Compounds .................................................................................. 2064 Factors That Affect Solubility ................................................................................................................. 2086 The Formation of Complex Ions ............................................................................................................. 2093 Solubility and pH...................................................................................................................................... 2111 Qualitative Analysis Using Selective Precipitation .............................................................................. 2130 End-of-Chapter Material ......................................................................................................................... 2135
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Chapter 18: Chemical Thermodynamics .................................................................... 2139 Thermodynamics and Work.................................................................................................................... 2141 The First Law of Thermodynamics......................................................................................................... 2154 The Second Law of Thermodynamics .................................................................................................... 2169 Entropy Changes and the Third Law of Thermodynamics .................................................................. 2190 Free Energy............................................................................................................................................... 2203 Spontaneity and Equilibrium.................................................................................................................. 2227 Comparing Thermodynamics and Kinetics ........................................................................................... 2243 Thermodynamics and Life....................................................................................................................... 2252 End-of-Chapter Material ......................................................................................................................... 2265
Chapter 19: Electrochemistry....................................................................................... 2272 Describing Electrochemical Cells ........................................................................................................... 2274 Standard Potentials ................................................................................................................................. 2291 Comparing Strengths of Oxidants and Reductants .............................................................................. 2321 Electrochemical Cells and Thermodynamics ........................................................................................ 2331 Commercial Galvanic Cells ...................................................................................................................... 2365 Corrosion .................................................................................................................................................. 2377 Electrolysis................................................................................................................................................ 2387 End-of-Chapter Material ......................................................................................................................... 2403
Chapter 20: Nuclear Chemistry .................................................................................... 2410 The Components of the Nucleus............................................................................................................. 2412 Nuclear Reactions .................................................................................................................................... 2425 The Interaction of Nuclear Radiation with Matter............................................................................... 2457 Thermodynamic Stability of the Atomic Nucleus ................................................................................ 2472 Applied Nuclear Chemistry ..................................................................................................................... 2493 The Origin of the Elements ..................................................................................................................... 2509 End-of-Chapter Material ......................................................................................................................... 2520
Chapter 21: Periodic Trends and the s-Block Elements ......................................... 2524 Overview of Periodic Trends................................................................................................................... 2526 The Chemistry of Hydrogen.................................................................................................................... 2538 The Alkali Metals (Group 1) .................................................................................................................... 2551 The Alkaline Earth Metals (Group 2)...................................................................................................... 2578 The s-Block Elements in Biology ............................................................................................................ 2597 End-of-Chapter Material ......................................................................................................................... 2607
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Chapter 22: The p-Block Elements .............................................................................. 2612 The Elements of Group 13 ....................................................................................................................... 2613 The Elements of Group 14 ....................................................................................................................... 2637 The Elements of Group 15 (The Pnicogens)........................................................................................... 2662 The Elements of Group 16 (The Chalcogens)......................................................................................... 2683 The Elements of Group 17 (The Halogens) ............................................................................................ 2702 The Elements of Group 18 (The Noble Gases)........................................................................................ 2717 End-of-Chapter Material ......................................................................................................................... 2730
Chapter 23: The d-Block Elements .............................................................................. 2735 General Trends among the Transition Metals ...................................................................................... 2736 A Brief Survey of Transition-Metal Chemistry ..................................................................................... 2748 Metallurgy ................................................................................................................................................ 2776 Coordination Compounds ....................................................................................................................... 2787 Crystal Field Theory................................................................................................................................. 2812 Transition Metals in Biology................................................................................................................... 2836 End-of-Chapter Material ......................................................................................................................... 2861
Chapter 24: Organic Compounds ................................................................................. 2865 Functional Groups and Classes of Organic Compounds ....................................................................... 2867 Isomers of Organic Compounds.............................................................................................................. 2872 Reactivity of Organic Molecules ............................................................................................................. 2897 Common Classes of Organic Reactions .................................................................................................. 2905 Common Classes of Organic Compounds............................................................................................... 2918 The Molecules of Life ............................................................................................................................... 2950 End-of-Chapter Material ......................................................................................................................... 2965
Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C .................................................................................................................................... 2970 Appendix B: Solubility-Product Constants (Ksp) for Compounds at 25°C .......... 2983 Appendix C: Dissociation Constants and pKa Values for Acids at 25°C............... 2989 Appendix D: Dissociation Constants and pKb Values for Bases at 25°C .............. 2992 Appendix E: Standard Reduction Potentials at 25°C ............................................... 2993 Appendix F: Properties of Water ................................................................................. 3000 Appendix G: Physical Constants and Conversion Factors ...................................... 3002 Appendix H: Periodic Table of Elements.................................................................... 3004 Appendix I: Experimentally Measured Masses of Selected Isotopes ................... 3010
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Art and Photo Credits .................................................................................................... 3012 Molecular Models..................................................................................................................................... 3013 Photo Credits ............................................................................................................................................ 3014
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About the Authors Bruce A. Averill Bruce A. Averill grew up in New England. He then received his B.S. with high honors in chemistry at Michigan State University in 1969, and his Ph.D. in inorganic chemistry at MIT in 1973. After three years as an NIH and NSF Postdoctoral Fellow at Brandeis University and the University of Wisconsin, he began his independent academic career at Michigan State University in 1976. He was promoted in 1982, after which he moved to the University of Virginia, where he was promoted to Professor in 1988. In 1994, Dr. Averill moved to the University of Amsterdam in the Netherlands as Professor of Biochemistry. He then returned to the United States to the University of Toledo in 2001, where he was a Distinguished University Professor. He was then named a Jefferson Science Policy Fellow at the U.S. State Department, where he remained for several years as a senior energy consultant. He is currently the founder and senior partner of Stategic Energy Security Solutions, which creates public/private partnerships to ensure global energy security. Dr. Averill’s academic research interests are centered on the role of metal ions in biology. He is also an expert on cyber-security. In his European position, Dr. Averill headed a European Union research network comprised of seven research groups from seven different European countries and a staff of approximately fifty research personnel. In addition, he was responsible for the research theme on Biocatalysis within the E. C. Slater Institute of the University of Amsterdam, which consisted of himself as head and a team of 21 professionals, ranging from associate professors to masters students at any given time. Dr. Averill’s research has attracted a great deal of attention in the scientific community. His published work is frequently cited by other researchers, and he has been invited to give more than 100 presentations at educational and research institutions and at national and international scientific meetings. Among his numerous awards, Dr. Averill has been an Honorary Woodrow Wilson Fellow, an NSF Predoctoral Fellow, an NIH and NSF Postdoctoral Fellow, and an Alfred P. Sloan Foundation Fellow; he has also received an NSF Special Creativity Award. Over the years, Dr. Averill has published more than 135 articles dealing with chemical, physical, and biological subjects in refereed journals, and he has also published 15 chapters in books and more than 80 abstracts from national and
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About the Authors
international meetings. In addition, he has co-edited a graduate text on catalysis, and he has taught courses at all levels, including general chemistry, biochemistry, advanced inorganic, and physical methods. Aside from his research program, Dr. Averill is an enthusiastic sailor and an avid reader. He also enjoys traveling with his family, and at some point in the future he would like to sail around the world in a classic wooden boat.
Patricia Eldredge Patricia Eldredge was raised in the U.S. diplomatic service, and has traveled and lived around the world. She has degrees from the Ohio State University, the University of Central Florida, the University of Virginia, and the University of North Carolina, Chapel Hill, where she obtained her Ph.D. in inorganic chemistry following several years as an analytical research chemist in industry. In addition, she has advanced offshore sailing qualifications from both the Royal Yachting Association in Britain and the American Sailing Association. In 1989, Dr. Eldredge was named the Science Policy Fellow for the American Chemical Society. While in Washington, D.C., she examined the impact of changes in federal funding priorities on academic research funding. She was awarded a Postdoctoral Research Fellowship with Oak Ridge Associated Universities, working with the U.S. Department of Energy on heterogeneous catalysis and coal liquefaction. Subsequently, she returned to the University of Virginia as a Research Scientist and a member of the General Faculty. In 1992, Dr. Eldredge relocated to Europe for several years. While there, she studied advanced Maritime Engineering, Materials, and Oceanography at the University of Southampton in England, arising from her keen interest in naval architecture. Upon her return to the United States in 2002, she was a Visiting Assistant Professor and a Senior Research Scientist at the University of Toledo. Her research interests included the use of protein scaffolds to synthesize biologically relevant clusters. Dr. Eldredge has published more than a dozen articles dealing with synthetic inorganic chemistry and catalysis, including several seminal studies describing new synthetic approaches to metal-sulfur clusters. She has also been awarded a patent for her work on catalytic coal liquefaction. Her diverse teaching experience includes courses on chemistry for the life sciences, introductory chemistry, general, organic, and analytical chemistry. When not
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About the Authors
authoring textbooks, Dr. Eldredge enjoys traveling, offshore sailing, political activism, and caring for her Havanese dogs.
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Acknowledgments The authors would like to thank the following individuals who reviewed the text and whose contributions were invaluable in shaping the product: • • • • • • • • • • • • • • • • • • • • • • • • •
Rebecca Barlag, Ohio University Greg Baxley, Cuesta College Karen Borgsmiller, Hood College Simon Bott, University of Houston David Burgess, Rivier College William Bushey, St. Marks High School and Delaware Technical Junior College Li-Heng Chen, Aquinas College Jose Conceicao, Metropolitan Community College Rajeev Dabke, Columbus State University Michael Denniston, Georgia Perimeter College Nathanael Fackler, Nebraska Wesleyan University James Fisher, Imperial Valley College Brian Gilbert, Linfield College Boyd Goodson, Southern Illinois University, Carbondale Karin Hassenrueck, California State University, Northridge James Hill, California State University, Sacramento Robert Holdar, North Lake College Roy Kennedy, Massachusetts Bay Community College Kristina Knutson, Georgia Perimeter College Chunmei Li, University of California, Berkeley Eric Malina, University of Nebraska Laura McCunn-Jordan, Marshall University Giovanni Meloni, University of San Francisco Mark Ott, Jackson Community College Robert Pike, The College of William & Mary
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Dedication To Harvey, who opened the door and to the Virginia Tech community for its resilience and strength. We Remember.
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Preface In this new millenium, as the world faces new and extreme challenges, the importance of acquiring a solid foundation in chemical principles has become increasingly important to understand the challenges that lie ahead. Moreover, as the world becomes more integrated and interdependent, so too do the scientific disciplines. The divisions between fields such as chemistry, physics, biology, environmental sciences, geology, and materials science, among others, have become less clearly defined. The goal of this text is to address the increasing close relationship among various disciplines and to show the relevance of chemistry to contemporary issues in a pedagogically approachable manner. Because of the enthusiasm of the majority of first-year chemistry students for biologically and medically relevant topics, this text uses an integrated approach that includes explicit discussions of biological and environmental applications of chemistry. Topics relevant to materials science are also introduced to meet the more specific needs of engineering students. To facilitate integration of such material, simple organic structures, nomenclature, and reactions are introduced very early in the text, and both organic and inorganic examples are used wherever possible. This approach emphasizes the distinctions between ionic and covalent bonding, thus enhancing the students’ chance of success in the organic chemistry course that traditionally follows general chemistry. The overall goal is to produce a text that introduces the students to the relevance and excitement of chemistry. Although much of first-year chemistry is taught as a service course, there is no reason that the intrinsic excitement and potential of chemistry cannot be the focal point of the text and the course. We emphasize the positive aspects of chemistry and its relationship to students’ lives, which requires bringing in applications early and often. Unfortunately, one cannot assume that students in such courses today are highly motivated to study chemistry for its own sake. The explicit discussion of biological, environmental, and materials issues from a chemical perspective is intended to motivate the students and help them appreciate the relevance of chemistry to their lives. Material that has traditionally been relegated to boxes, and thus perhaps perceived as peripheral by the students, has been incorporated into the text to serve as a learning tool. To begin the discussion of chemistry rapidly, the traditional first chapter introducing units, significant figures, conversion factors, dimensional analysis, and so on, has been reorganized. The material has been placed in the chapters where the relevant concepts are first introduced, thus providing three advantages: it
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Preface
eliminates the tedium of the traditional approach, which introduces mathematical operations at the outset, and thus avoids the perception that chemistry is a mathematics course; it avoids the early introduction of operations such as logarithms and exponents, which are typically not encountered again for several chapters and may easily be forgotten when they are needed; and third, it provides a review for those students who have already had relatively sophisticated high school chemistry and math courses, although the sections are designed primarily for students unfamiliar with the topic. Our specific objectives include the following: 1. To write the text at a level suitable for science majors, but using a less formal writing style that will appeal to modern students. 2. To produce a truly integrated text that gives the student who takes only a single year of chemistry an overview of the most important subdisciplines of chemistry, including organic, inorganic, biological, materials, environmental, and nuclear chemistry, thus emphasizing unifying concepts. 3. To introduce fundamental concepts in the first two-thirds of the chapter, then applications relevant to the health sciences or engineers. This provides a flexible text that can be tailored to the specific needs and interests of the audience. 4. To ensure the accuracy of the material presented, which is enhanced by the author’s breadth of professional experience and experience as active chemical researchers. 5. To produce a spare, clean, uncluttered text that is less distracting to the student, where each piece of art serves as a pedagogical device. 6. To introduce the distinction between ionic and covalent bonding and reactions early in the text, and to continue to build on this foundation in the subsequent discussion, while emphasizing the relationship between structure and reactivity. 7. To utilize established pedagogical devices to maximize students’ ability to learn directly from the text. These include copious worked examples in the text, problem-solving strategies, and similar unworked exercises with solutions. End-of-chapter problems are designed to ensure that students have grasped major concepts in addition to testing their ability to solve numerical, problems. Problems emphasizing applications are drawn from many disciplines. 8. To emphasize an intuitive and predictive approach to problem solving that relies on a thorough understanding of key concepts and recognition of important patterns rather than on memorization. Many patterns are indicated throughout the text as notes in the margin.
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Preface
The text is organized by units that discuss introductory concepts, atomic and molecular structure, the states of matter, kinetics and equilibria, and descriptive inorganic chemistry. The text breaks the traditional chapter on liquids and solids into two to expand the coverage of important and topics such as semiconductors and superconductors, polymers, and engineering materials. In summary, this text represents a step in the evolution of the general chemistry text toward one that reflects the increasing overlap between chemistry and other disciplines. Most importantly, the text discusses exciting and relevant aspects of biological, environmental, and materials science that are usually relegated to the last few chapters, and it provides a format that allows the instructor to tailor the emphasis to the needs of the class. By the end of Chapter 6 "The Structure of Atoms", the student will have already been introduced to environmental topics such as acid rain, the ozone layer, and periodic extinctions, and to biological topics such as antibiotics and the caloric content of foods. Nonetheless, the new material is presented in such a way as to minimally perturb the traditional sequence of topics in a first-year course, making the adaptation easier for instructors.
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Chapter 1 Introduction to Chemistry As you begin your study of college chemistry, those of you who do not intend to become professional chemists may well wonder why you need to study chemistry. You will soon discover that a basic understanding of chemistry is useful in a wide range of disciplines and career paths. You will also discover that an understanding of chemistry helps you make informed decisions about many issues that affect you, your community, and your world. A major goal of this text is to demonstrate the importance of chemistry in your daily life and in our collective understanding of both the physical world we occupy and the biological realm of which we are a part. The objectives of this chapter are twofold: (1) to introduce the breadth, the importance, and some of the challenges of modern chemistry and (2) to present some of the fundamental concepts and definitions you will need to understand how chemists think and work.
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Chapter 1 Introduction to Chemistry
An atomic corral for electrons. A corral of 48 iron atoms (yellow-orange) on a smooth copper surface (cyanpurple) confines the electrons on the surface of the copper, producing a pattern of “ripples” in the distribution of the electrons. Scientists assembled the 713-picometer-diameter corral by individually positioning iron atoms with the tip of a scanning tunneling microscope. (Note that 1 picometer is equivalent to 1 × 10-12 meters.)
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1.1 Chemistry in the Modern World LEARNING OBJECTIVE 1. To recognize the breadth, depth, and scope of chemistry.
Chemistry1 is the study of matter and the changes that material substances undergo. Of all the scientific disciplines, it is perhaps the most extensively connected to other fields of study. Geologists who want to locate new mineral or oil deposits use chemical techniques to analyze and identify rock samples. Oceanographers use chemistry to track ocean currents, determine the flux of nutrients into the sea, and measure the rate of exchange of nutrients between ocean layers. Engineers consider the relationships between the structures and the properties of substances when they specify materials for various uses. Physicists take advantage of the properties of substances to detect new subatomic particles. Astronomers use chemical signatures to determine the age and distance of stars and thus answer questions about how stars form and how old the universe is. The entire subject of environmental science depends on chemistry to explain the origin and impacts of phenomena such as air pollution, ozone layer depletion, and global warming. The disciplines that focus on living organisms and their interactions with the physical world rely heavily on biochemistry2, the application of chemistry to the study of biological processes. A living cell contains a large collection of complex molecules that carry out thousands of chemical reactions, including those that are necessary for the cell to reproduce. Biological phenomena such as vision, taste, smell, and movement result from numerous chemical reactions. Fields such as medicine, pharmacology, nutrition, and toxicology focus specifically on how the chemical substances that enter our bodies interact with the chemical components of the body to maintain our health and well-being. For example, in the specialized area of sports medicine, a knowledge of chemistry is needed to understand why muscles get sore after exercise as well as how prolonged exercise produces the euphoric feeling known as “runner’s high.” 1. The study of matter and the changes that material substances undergo. 2. The application of chemistry to the study of biological processes.
Examples of the practical applications of chemistry are everywhere (Figure 1.1 "Chemistry in Everyday Life"). Engineers need to understand the chemical properties of the substances when designing biologically compatible implants for joint replacements or designing roads, bridges, buildings, and nuclear reactors that do not collapse because of weakened structural materials such as steel and cement.
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Archaeology and paleontology rely on chemical techniques to date bones and artifacts and identify their origins. Although law is not normally considered a field related to chemistry, forensic scientists use chemical methods to analyze blood, fibers, and other evidence as they investigate crimes. In particular, DNA matching—comparing biological samples of genetic material to see whether they could have come from the same person—has been used to solve many high-profile criminal cases as well as clear innocent people who have been wrongly accused or convicted. Forensics is a rapidly growing area of applied chemistry. In addition, the proliferation of chemical and biochemical innovations in industry is producing rapid growth in the area of patent law. Ultimately, the dispersal of information in all the fields in which chemistry plays a part requires experts who are able to explain complex chemical issues to the public through television, print journalism, the Internet, and popular books. Figure 1.1 Chemistry in Everyday Life
Although most people do not recognize it, chemistry and chemical compounds are crucial ingredients in almost everything we eat, wear, and use.
1.1 Chemistry in the Modern World
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Chapter 1 Introduction to Chemistry
By this point, it shouldn’t surprise you to learn that chemistry was essential in explaining a pivotal event in the history of Earth: the disappearance of the dinosaurs. Although dinosaurs ruled Earth for more than 150 million years, fossil evidence suggests that they became extinct rather abruptly approximately 66 million years ago. Proposed explanations for their extinction have ranged from an epidemic caused by some deadly microbe or virus to more gradual phenomena such as massive climate changes. In 1978 Luis Alvarez (a Nobel Prize–winning physicist), the geologist Walter Alvarez (Luis’s son), and their coworkers discovered a thin layer of sedimentary rock formed 66 million years ago that contained unusually high concentrations of iridium, a rather rare metal (part (a) in Figure 1.2 "Evidence for the Asteroid Impact That May Have Caused the Extinction of the Dinosaurs"). This layer was deposited at about the time dinosaurs disappeared from the fossil record. Although iridium is very rare in most rocks, accounting for only 0.0000001% of Earth’s crust, it is much more abundant in comets and asteroids. Because corresponding samples of rocks at sites in Italy and Denmark contained high iridium concentrations, the Alvarezes suggested that the impact of a large asteroid with Earth led to the extinction of the dinosaurs. When chemists analyzed additional samples of 66-million-year-old sediments from sites around the world, all were found to contain high levels of iridium. In addition, small grains of quartz in most of the iridium-containing layers exhibit microscopic cracks characteristic of high-intensity shock waves (part (b) in Figure 1.2 "Evidence for the Asteroid Impact That May Have Caused the Extinction of the Dinosaurs"). These grains apparently originated from terrestrial rocks at the impact site, which were pulverized on impact and blasted into the upper atmosphere before they settled out all over the world.
1.1 Chemistry in the Modern World
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Chapter 1 Introduction to Chemistry
Scientists calculate that a collision of Earth with a stony asteroid about 10 kilometers (6 miles) in diameter, traveling at 25 kilometers per second (about 56,000 miles per hour), would almost instantaneously release energy equivalent to the explosion of about 100 million megatons of TNT (trinitrotoluene). This is more energy than that stored in the entire nuclear arsenal of the world. The energy released by such an impact would set fire to vast areas of forest, and the smoke from the fires and the dust created by the impact would block the sunlight for months or years, eventually killing virtually all green plants and most organisms that depend on them. This could explain why about 70% of all species—not just dinosaurs—disappeared at the same time. Scientists also calculate that this impact would form a crater at least 125 kilometers (78 miles) in diameter. Recently, satellite images from a Space Shuttle mission confirmed that a huge asteroid or comet crashed into Earth’s surface across the Yucatan’s northern tip in the Gulf of Mexico 65 million years ago, leaving a partially submerged crater 180 kilometers (112 miles) in diameter (Figure 1.3 "Asteroid Impact"). Thus simple chemical measurements of the abundance of one element in rocks led to a new and dramatic explanation for the extinction of the dinosaurs. Though still controversial, this explanation is supported by additional evidence, much of it chemical.
1.1 Chemistry in the Modern World
Figure 1.2 Evidence for the Asteroid Impact That May Have Caused the Extinction of the Dinosaurs
(a) Luis and Walter Alvarez are standing in front of a rock formation in Italy that shows the thin white layer of iridium-rich clay deposited at the time the dinosaurs became extinct. The concentration of iridium is 30 times higher in this layer than in the rocks immediately above and below it. There are no significant differences between the clay layer and the surrounding rocks in the concentrations of any of the 28 other elements examined. (b) Microphotographs of an unshocked quartz grain (left) and a quartz grain from the iridiumrich layer exhibiting microscopic cracks resulting from shock (right).
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Figure 1.3 Asteroid Impact
The location of the asteroid impact crater near what is now the tip of the Yucatan Peninsula in Mexico.
This is only one example of how chemistry has been applied to an important scientific problem. Other chemical applications and explanations that we will discuss in this text include how astronomers determine the distance of galaxies and how fish can survive in subfreezing water under polar ice sheets. We will also consider ways in which chemistry affects our daily lives: the addition of iodine to table salt; the development of more effective drugs to treat diseases such as cancer, AIDS (acquired immunodeficiency syndrome), and arthritis; the retooling of industry to use nonchlorine-containing refrigerants, propellants, and other chemicals to preserve Earth’s ozone layer; the use of modern materials in engineering; current efforts to control the problems of acid rain and global warming; and the awareness that our bodies require small amounts of some chemical substances that are toxic when ingested in larger doses. By the time you finish this text, you will be able to discuss these kinds of topics knowledgeably, either as a beginning scientist who intends to spend your career studying such problems or as an informed observer who is able to participate in public debates that will certainly arise as society grapples with scientific issues.
1.1 Chemistry in the Modern World
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Summary Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine.
KEY TAKEAWAY • An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines.
1.1 Chemistry in the Modern World
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Chapter 1 Introduction to Chemistry
1.2 The Scientific Method LEARNING OBJECTIVE 1. To identify the components of the scientific method.
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method3. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure 1.4 "The Scientific Method"). Figure 1.4 The Scientific Method
3. The procedure that scientists use to search for answers to questions and solutions to problems.
As depicted in this flowchart, the scientific method consists of making observations, formulating hypotheses, and designing experiments. A scientist may enter the cycle at any point.
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Chapter 1 Introduction to Chemistry
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 degrees Celsius, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 degrees Celsius. For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal. After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis4, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses: (1) Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or (2) the sun revolves around Earth every 24 hours. Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either support or refute it. 4. A tentative explanation for scientific observations that puts the system being studied into a form that can be tested. 5. A systematic observation or measurement, preferably made under controlled conditions—that is, conditions in which the variable of interest is clearly distinguished from any others. 6. A verbal or mathematical description of a phenomenon that allows for general predictions and says what happens, not why it happens. 7. A chemical substance always contains the same proportions of elements by mass.
1.2 The Scientific Method
After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments5 are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in our extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law6, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why. One example of a law, the law of definite proportions7, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table
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Chapter 1 Introduction to Chemistry
salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass.You will learn in Chapter 12 "Solids" that some solid compounds do not strictly obey the law of definite proportions. (For a review of common units of measurement, see Essential Skills 1 in Section 1.9 "Essential Skills 1".) The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century. Whereas a law states only what happens, a theory8 attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. Figure 1.5 "A Summary of How the Scientific Method Was Used in Developing the Asteroid Impact Theory to Explain the Disappearance of the Dinosaurs from Earth" summarizes the application of the scientific method in this case.
8. A statement that attempts to explain why nature behaves the way it does.
1.2 The Scientific Method
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Chapter 1 Introduction to Chemistry
Figure 1.5 A Summary of How the Scientific Method Was Used in Developing the Asteroid Impact Theory to Explain the Disappearance of the Dinosaurs from Earth
1.2 The Scientific Method
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Chapter 1 Introduction to Chemistry
EXAMPLE 1 Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. a. Ice always floats on liquid water. b. Birds evolved from dinosaurs. c. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly. d. When 10 g of ice were added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted. e. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised. Given: components of the scientific method Asked for: statement classification Strategy: Refer to the definitions in this section to determine which category best describes each statement. Solution: a. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law. b. This is a possible explanation for the origin of birds, so it is a hypothesis. c. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory. d. The temperature is measured before and after a change is made in a system, so these are quantitative observations. e. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment. Exercise Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1.2 The Scientific Method
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Chapter 1 Introduction to Chemistry
a. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.” b. Heat always flows from hot objects to cooler ones, not in the opposite direction. c. The universe was formed by a massive explosion that propelled matter into a vacuum. d. Michael Jordan is the greatest pure shooter ever to play professional basketball. e. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas. f. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive. Answer: a. b. c. d. e. f.
experiment law theory hypothesis qualitative observation quantitative observation
Because scientists can enter the cycle shown in Figure 1.4 "The Scientific Method" at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations. It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology.
1.2 The Scientific Method
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Chapter 1 Introduction to Chemistry
In Section 1.3 "A Description of Matter", we begin our discussion of chemistry with a description of matter. This discussion is followed by a summary of some of the pioneering discoveries that led to our present understanding of the structure of the fundamental unit of chemistry: the atom.
Summary Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method.
KEY TAKEAWAY • Chemists expand their knowledge with the scientific method.
1.2 The Scientific Method
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CONCEPTUAL PROBLEMS 1. What are the three components of the scientific method? Is it necessary for an individual to conduct experiments to follow the scientific method? 2. Identify each statement as a theory or a law and explain your reasoning. a. The ratio of elements in a pure substance is constant. b. An object appears black because it absorbs all the visible light that strikes it. c. Energy is neither created nor destroyed. d. Metals conduct electricity because their electrons are not tightly bound to a particular nucleus and are therefore free to migrate. 3. Identify each statement as a theory or a law and explain your reasoning. a. A pure chemical substance contains the same proportion of elements by mass. b. The universe is expanding. c. Oppositely charged particles attract each other. d. Life exists on other planets. 4. Classify each statement as a qualitative observation or a quantitative observation. a. Mercury and bromine are the only elements that are liquids at room temperature. b. An element is both malleable and ductile. c. The density of iron is 7.87 g/cm3. d. Lead absorbs sound very effectively. e. A meteorite contains 20% nickel by mass. 5. Classify each statement as a quantitative observation or a qualitative observation. a. b. c. d.
1.2 The Scientific Method
Nickel deficiency in rats is associated with retarded growth. Boron is a good conductor of electricity at high temperatures. There are 1.4–2.3 g of zinc in an average 70 kg adult. Certain osmium compounds found in air in concentrations as low as 10.7 µg/m3 can cause lung cancer.
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ANSWERS
1.2 The Scientific Method
3. a. b. c. d.
law theory law theory
5. a. b. c. d.
qualitative qualitative quantitative quantitative
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Chapter 1 Introduction to Chemistry
1.3 A Description of Matter LEARNING OBJECTIVE 1. To classify matter.
Chemists study the structures, physical properties, and chemical properties of material substances. These consist of matter9, which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter. 9. Anything that occupies space and has mass. 10. A fundamental property that does not depend on an object’s location; it is the quantity of matter an object contains. 11. A force caused by the gravitational attraction that operates on an object. The weight of an object depends on its location (c.f. mass). 12. One of three distinct states of matter that, under normal conditions, is relatively rigid and has a fixed volume. 13. One of three distinct states of matter that, under normal conditons, has a fixed volume but flows to assume the shape of its container. 14. One of three distinct states of matter that, under normal conditions, has neither a fixed shape nor a fixed volume and expands to completely fill its container. 15. The amount of force exerted on a given area. 16. A change of state that does not affect the chemical composition of a substance.
The mass10 of an object is the quantity of matter it contains. Do not confuse an object’s mass with its weight11, which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 (Section 1.9 "Essential Skills 1"). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth. Under normal conditions, there are three distinct states of matter: solids, liquids, and gases (Figure 1.6 "The Three States of Matter"). Solids12 are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, liquids13 have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. Gases14, such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and pressure15 (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a physical change16. For example, liquid water can be heated to form a gas called steam, or steam can be cooled to
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form liquid water. However, such changes of state do not affect the chemical composition of the substance. Figure 1.6 The Three States of Matter
Solids have a defined shape and volume. Liquids have a fixed volume but flow to assume the shape of their containers. Gases completely fill their containers, regardless of volume.
Pure Substances and Mixtures
17. A combination of two or more pure substances in variable proportions in which the individual substances retain their respective identities. 18. A mixture in which all portions of a material are in the same state, have no visible boundaries, and are uniform throughout.
1.3 A Description of Matter
A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixtures17, which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is homogeneous18. Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions
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of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys. If the composition of a material is not completely uniform, then it is heterogeneous19 (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water (Figure 1.7 "A Heterogeneous Mixture"). The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods.
19. A mixture in which a material is not completely uniform throughout. 20. A physical process used to separate homogeneous mixtures (solutions) into their component substances. Distillation makes use of differences in the volatilities of the component substances.
1.3 A Description of Matter
Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. Distillation20 makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. Figure 1.8 "The Distillation of a Solution of Table Salt in Water" shows a simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.
Figure 1.7 A Heterogeneous Mixture
Under a microscope, whole milk is actually a heterogeneous mixture composed of globules of fat and protein dispersed in water.
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Figure 1.8 The Distillation of a Solution of Table Salt in Water
The solution of salt in water is heated in the distilling flask until it boils. The resulting vapor is enriched in the more volatile component (water), which condenses to a liquid in the cold condenser and is then collected in the receiving flask.
Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States.
21. A physical process used to separate homogeneous mixtures (solutions) into their component substances. Crystallization separates mixtures based on differences in their solubilities.
1.3 A Description of Matter
Crystallization21 separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals
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can be removed from the remaining solution by filtration. Figure 1.9 "The Crystallization of Sodium Acetate from a Concentrated Solution of Sodium Acetate in Water" dramatically illustrates the process of crystallization. Figure 1.9 The Crystallization of Sodium Acetate from a Concentrated Solution of Sodium Acetate in Water
The addition of a small “seed” crystal (a) causes the compound to form white crystals, which grow and eventually occupy most of the flask (b).
22. A pure substance that cannot be broken down into a simpler substance by chemical changes. 23. A pure substance that contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. 24. A process in which the chemical composition of one or more substances is altered.
1.3 A Description of Matter
Most mixtures can be separated into pure substances, which may be either elements or compounds. An element22, such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a compound23, such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a chemical change24. The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 115 elements are known, but millions of chemical compounds have
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been prepared from these 115 elements. The known elements are listed in the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"). In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process called electrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements (Figure 1.10 "The Decomposition of Water to Hydrogen and Oxygen by Electrolysis"). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both costeffective and ecologically sound. Figure 1.10 The Decomposition of Water to Hydrogen and Oxygen by Electrolysis
Water is a chemical compound; hydrogen and oxygen are elements.
The overall organization of matter and the methods used to separate mixtures are summarized in Figure 1.11 "Relationships between the Types of Matter and the Methods Used to Separate Mixtures".
1.3 A Description of Matter
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Figure 1.11 Relationships between the Types of Matter and the Methods Used to Separate Mixtures
1.3 A Description of Matter
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EXAMPLE 2 Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution). a. b. c. d.
filtered tea freshly squeezed orange juice a compact disc aluminum oxide, a white powder that contains a 2:3 ratio of aluminum and oxygen atoms e. selenium
Given: a chemical substance Asked for: its classification Strategy: A Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound. B If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture. Solution: a. A Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration. B Because the composition of the solution is uniform throughout, it is a homogeneous mixture. b. A Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure. B Because its composition is not uniform throughout, orange juice is a heterogeneous mixture. c. A A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence a compact disc is not chemically pure. B The regions of different composition indicate that a compact disc is a heterogeneous mixture. d. A Aluminum oxide is a single, chemically pure compound. e. A Selenium is one of the known elements.
1.3 A Description of Matter
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Chapter 1 Introduction to Chemistry
Exercise Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution). a. b. c. d.
white wine mercury ranch-style salad dressing table sugar (sucrose)
Answer: a. b. c. d.
solution element heterogeneous mixture compound
Properties of Matter All matter has physical and chemical properties. Physical properties25 are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties26 describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property). 25. A characteristic that scientists can measure without changing the composition of a sample under study. 26. The characteristic ability of a substance to react to form new substances. 27. A physical property that varies with the amount of a substance. 28. A physical property that does not depend on the amount of the substance and physical state at a given temperature and pressure.
1.3 A Description of Matter
Physical properties can be extensive or intensive. Extensive properties27 vary with the amount of the substance and include mass, weight, and volume. Intensive properties28, in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature. For example, elemental sulfur is a yellow crystalline solid that does not conduct electricity and has a melting point of 115.2°C, no matter what amount is examined (Figure 1.12 "The Difference between Extensive and Intensive Properties of Matter"). Scientists commonly measure intensive properties to determine a substance’s identity, whereas extensive properties convey information about the amount of the substance in a sample.
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Figure 1.12 The Difference between Extensive and Intensive Properties of Matter
Because they differ in size, the two samples of sulfur have different extensive properties, such as mass and volume. In contrast, their intensive properties, including color, melting point, and electrical conductivity, are identical.
Although mass and volume are both extensive properties, their ratio is an important intensive property called density (d)29. Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm 3). As mass increases in a given volume, density also increases. For example, lead, with its greater mass, has a far greater density than the same volume of air, just as a brick has a greater density than the same volume of Styrofoam. At a given temperature and pressure, the density of a pure substance is a constant: Equation 1.1
density =
mass m ⇒d= volume v
Pure water, for example, has a density of 0.998 g/cm3 at 25°C. The average densities of some common substances are in Table 1.1 "Densities of Common Substances". Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.” Example 3 shows how density measurements can be used to identify pure substances. Table 1.1 Densities of Common Substances 29. An intensive property of matter, density is the mass per unit volume (usually expressed in g/cm3). At a given temperature, the density of a substance is a constant.
1.3 A Description of Matter
Substance
Density at 25°C (g/cm3)
blood
1.035
body fat
0.918
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1.3 A Description of Matter
Substance
Density at 25°C (g/cm3)
whole milk
1.030
corn oil
0.922
mayonnaise
0.910
honey
1.420
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EXAMPLE 3 The densities of some common liquids are in Table 1.2 "Densities of Liquids in Example 3". Imagine you have five bottles containing colorless liquids (labeled A–E). You must identify them by measuring the density of each. Using a pipette, a laboratory instrument for accurately measuring and transferring liquids, you carefully measure 25.00 mL of each liquid into five beakers of known mass (1 mL = 1 cm3). You then weigh each sample on a laboratory balance. Use the tabulated data to calculate the density of each sample. Based solely on your results, can you unambiguously identify all five liquids?If necessary, review the use of significant figures in calculations in Essential Skills 1 (Section 1.9 "Essential Skills 1") prior to working this example. Masses of samples: A, 17.72 g; B, 19.75 g; C, 24.91 g; D, 19.65 g; E, 27.80 g
TABLE 1.2 DENSITIES OF LIQUIDS IN EXAMPLE 3 Substance
Density at 25°C (g/cm3)
water
0.998
ethanol (the alcohol in beverages)
0.789
methanol (wood alcohol)
0.792
ethylene glycol (used in antifreeze)
1.113
diethyl ether (“ether”; once widely used as an anesthetic)
0.708
isopropanol (rubbing alcohol)
0.785
Given: volume and mass Asked for: density Strategy: A Calculate the density of each liquid from the volumes and masses given. B Check to make sure that your answer makes sense.
1.3 A Description of Matter
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C Compare each calculated density with those given in Table 1.2 "Densities of Liquids in Example 3". If the calculated density of a liquid is not significantly different from that of one of the liquids given in the table, then the unknown liquid is most likely the corresponding liquid. D If none of the reported densities corresponds to the calculated density, then the liquid cannot be unambiguously identified. Solution: A Density is mass per unit volume and is usually reported in grams per cubic centimeter (or grams per milliliter because 1 mL = 1 cm 3). The masses of the samples are given in grams, and the volume of all the samples is 25.00 mL (= 25.00 cm3). The density of each sample is calculated by dividing the mass by its volume (Equation 1.1). The density of sample A, for example, is
17. 72 g = 0.7088 g/cm 3 3 25.00 cm Both the volume and the mass are given to four significant figures, so four significant figures are permitted in the result. (See Essential Skills 1, Section 1.9 "Essential Skills 1", for a discussion of significant figures.) The densities of the other samples (in grams per cubic centimeter) are as follows: B, 0.7900; C, 0.9964; D, 0.7860; and E, 1.112. B Except for sample E, the calculated densities are slightly less than 1 g/cm 3. This makes sense because the masses (in grams) of samples A–D are all slightly less than the volume of the samples, 25.00 mL. In contrast, the mass of sample E is slightly greater than 25 g, so its density must be somewhat greater than 1 g/cm3. C Comparing these results with the data given in Table 1.2 "Densities of Liquids in Example 3" shows that sample A is probably diethyl ether (0.708 g/cm3 and 0.7088 g/cm3 are not substantially different), sample C is probably water (0.998 g/cm3 in the table versus 0.9964 g/cm3 measured), and sample E is probably ethylene glycol (1.113 g/cm3 in the table versus 1.112 g/cm3 measured). D Samples B and D are more difficult to identify for two reasons: (1) Both have similar densities (0.7900 and 0.7860 g/cm3), so they may or may not be chemically identical. (2) Within experimental error, the measured densities of B and D are indistinguishable from the densities of ethanol (0.789 g/cm 3),
1.3 A Description of Matter
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methanol (0.792 g/cm3), and isopropanol (0.785 g/cm3). Thus some property other than density must be used to identify each sample. Exercise Given the volumes and masses of five samples of compounds used in blending gasoline, together with the densities of several chemically pure liquids, identify as many of the samples as possible. Sample Volume (mL) Mass (g) A
337
250.0
B
972
678.1
C
243
190.9
D
119
103.2
E
499
438.7
Substance
Density (g/cm3)
benzene
0.8787
toluene
0.8669
m-xylene
0.8684
isooctane
0.6979
methyl t-butyl ether
0.7405
t-butyl alcohol
0.7856
Answer: A, methyl t-butyl ether; B, isooctane; C, t-butyl alcohol; D, toluene or m-xylene; E, benzene
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Summary Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density.
KEY TAKEAWAY • Matter can be classified according to physical and chemical properties.
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CONCEPTUAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Conceptual Problems. 1. What is the difference between mass and weight? Is the mass of an object on Earth the same as the mass of the same object on Jupiter? Why or why not? 2. Is it accurate to say that a substance with a mass of 1 kg weighs 2.2 lb? Why or why not? 3. What factor must be considered when reporting the weight of an object as opposed to its mass? 4. Construct a table with the headings “Solid,” “Liquid,” and “Gas.” For any given substance, state what you expect for each of the following: a. b. c. d. e.
the relative densities of the three phases the physical shapes of the three phases the volumes for the same mass of compound the sensitivity of the volume of each phase to changes in temperature the sensitivity of the volume to changes in pressure
5. Classify each substance as homogeneous or heterogeneous and explain your reasoning. a. b. c. d. e. f.
platinum a carbonated beverage bronze wood natural gas Styrofoam
6. Classify each substance as homogeneous or heterogeneous and explain your reasoning. a. b. c. d. e. f.
snowflakes gasoline black tea plastic wrap blood water containing ice cubes
7. Classify each substance as a pure substance or a mixture and explain your reasoning. a. seawater
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b. c. d. e.
coffee 14-karat gold diamond distilled water
8. Classify each substance as a pure substance or a mixture. a. b. c. d. e.
cardboard caffeine tin a vitamin tablet helium gas
9. Classify each substance as an element or a compound. a. b. c. d. e.
sugar silver rust rubbing alcohol copper
10. Classify each substance as an element or a compound. a. b. c. d. e.
water iron hydrogen gas glass nylon
11. What techniques could be used to separate each of the following? a. sugar and water from an aqueous solution of sugar b. a mixture of sugar and sand c. a heterogeneous mixture of solids with different solubilities 12. What techniques could be used to separate each of the following? a. solid calcium chloride from a solution of calcium chloride in water b. the components of a solution of vinegar in water c. particulates from water in a fish tank 13. Match each separation technique in (a) with the physical/chemical property that each takes advantage of in (b). a. crystallization, distillation, filtration b. volatility, physical state, solubility
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14. The following figures illustrate the arrangement of atoms in some samples of matter. Which figures are related by a physical change? By a chemical change?
15. Classify each statement as an extensive property or an intensive property. a. b. c. d. e.
Carbon, in the form of diamond, is one of the hardest known materials. A sample of crystalline silicon, a grayish solid, has a mass of 14.3 g. Germanium has a density of 5.32 g/cm3. Gray tin converts to white tin at 13.2°C. Lead is a bluish-white metal.
16. Classify each statement as a physical property or a chemical property. a. Fluorine etches glass. b. Chlorine interacts with moisture in the lungs to produce a respiratory irritant. c. Bromine is a reddish-brown liquid. d. Iodine has a density of 11.27 g/L at 0°C.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Numerical Problems. 1. If a person weighs 176 lb on Earth, what is his or her mass on Mars, where the force of gravity is 37% of that on Earth? 2. If a person weighs 135 lb on Earth, what is his or her mass on Jupiter, where the force of gravity is 236% of that on Earth? 3. Calculate the volume of 10.00 g of each element and then arrange the elements in order of decreasing volume. The numbers in parentheses are densities. a. b. c. d.
copper (8.92 g/cm3) calcium (1.54 g/cm3) titanium (4.51 g/cm3) iridium (22.85 g/cm3)
4. Given 15.00 g of each element, calculate the volume of each and then arrange the elements in order of increasing volume. The numbers in parentheses are densities. a. b. c. d.
gold (19.32 g/cm3) lead (11.34 g/cm3) iron (7.87 g/cm3) sulfur (2.07 g/cm3)
5. A silver bar has dimensions of 10.00 cm × 4.00 cm × 1.50 cm, and the density of silver is 10.49 g/cm3. What is the mass of the bar? 6. Platinum has a density of 21.45 g/cm3. What is the mass of a platinum bar measuring 3.00 cm × 1.50 cm × 0.500 cm? 7. Complete the following table. Density (g/cm3) Mass (g) Volume (cm3) Element 3.14
79.904
3.51 39.1 11.34
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3.42
C
45.5
K
207.2 107.868
6.51
Br
Pb 10.28
Ag
14.0
Zr
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8. Gold has a density of 19.30 g/cm3. If a person who weighs 85.00 kg (1 kg = 1000 g) were given his or her weight in gold, what volume (in cm 3) would the gold occupy? Are we justified in using the SI unit of mass for the person’s weight in this case? 9. An irregularly shaped piece of magnesium with a mass of 11.81 g was dropped into a graduated cylinder partially filled with water. The magnesium displaced 6.80 mL of water. What is the density of magnesium? 10. The density of copper is 8.92 g/cm3. If a 10.00 g sample is placed in a graduated cylinder that contains 15.0 mL of water, what is the total volume that would be occupied? 11. At 20°C, the density of fresh water is 0.9982 kg/m3, and the density of seawater is 1.025 kg/m3. Will a ship float higher in fresh water or in seawater? Explain your reasoning.
ANSWERS 1. Unlike weight, mass does not depend on location. The mass of the person is therefore the same on Earth and Mars: 176 lb ÷ 2.2 lb/kg = 80 kg. 3. a. b. c. d.
Cu: 1.12 cm3 Ca: 6.49 cm3 Ti: 2.22 cm3 Ir: 0.4376 cm3
Volume decreases: Ca > Ti > Cu > Ir 5. 629 g 9. 1.74 g/cm3
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1.4 A Brief History of Chemistry LEARNING OBJECTIVE 1. To understand the development of the atomic model.
It was not until the era of the ancient Greeks that we have any record of how people tried to explain the chemical changes they observed and used. At that time, natural objects were thought to consist of only four basic elements: earth, air, fire, and water. Then, in the fourth century BC, two Greek philosophers, Democritus and Leucippus, suggested that matter was not infinitely divisible into smaller particles but instead consisted of fundamental, indivisible particles called atoms30. Unfortunately, these early philosophers did not have the technology to test their hypothesis. They would have been unlikely to do so in any case because the ancient Greeks did not conduct experiments or use the scientific method. They believed that the nature of the universe could be discovered by rational thought alone. Over the next two millennia, alchemists, who engaged in a form of chemistry and speculative philosophy during the Middle Ages and Renaissance, achieved many advances in chemistry. Their major goal was to convert certain elements into others by a process they called transmutation31 (Figure 1.13 "An Alchemist at Work"). In particular, alchemists wanted to find a way to transform cheaper metals into gold. Although most alchemists did not approach chemistry systematically and many appear to have been outright frauds, alchemists in China, the Arab kingdoms, and medieval Europe made major contributions, including the discovery of elements such as quicksilver (mercury) and the preparation of several strong acids.
30. The fundamental, individual particles of which matter is composed. 31. The process of converting one element to another.
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Modern Chemistry Figure 1.13 An Alchemist at
The 16th and 17th centuries saw the beginnings of what Work we now recognize as modern chemistry. During this period, great advances were made in metallurgy, the extraction of metals from ores, and the first systematic quantitative experiments were carried out. In 1661, the Englishman Robert Boyle (1627–91) published The Sceptical Chymist, which described the relationship between the pressure and the volume of air. More important, Boyle defined an element as a substance that cannot be broken down into two or more simpler substances by chemical means. This led to the identification of a large number of elements, many of Alchemy was a form of chemistry that flourished during the Middle which were metals. Ironically, Boyle himself never Ages and Renaissance. Although thought that metals were elements.
some alchemists were frauds, others made major contributions, including the discovery of several elements and the preparation of strong acids.
In the 18th century, the English clergyman Joseph Priestley (1733–1804) discovered oxygen gas and found that many carbon-containing materials burn vigorously in an oxygen atmosphere, a process called combustion32. Priestley also discovered that the gas produced by fermenting beer, which we now know to be carbon dioxide, is the same as one of the gaseous products of combustion. Priestley’s studies of this gas did not continue as he would have liked, however. After he fell into a vat of fermenting beer, brewers prohibited him from working in their factories. Although Priestley did not understand its identity, he found that carbon dioxide dissolved in water to produce seltzer water. In essence, he may be considered the founder of the multibillion-dollar carbonated soft drink industry.
32. The burning of a material in an oxygen atmosphere.
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Joseph Priestley (1733–1804) Priestley was a political theorist and a leading Unitarian minister. He was appointed to Warrington Academy in Lancashire, England, where he developed new courses on history, science, and the arts. During visits to London, Priestley met the leading men of science, including Benjamin Franklin, who encouraged Priestley’s interest in electricity. Priestley’s work on gases began while he was living next to a brewery in Leeds, where he noticed “fixed air” bubbling out of vats of fermenting beer and ale. His scientific discoveries included the relationship between electricity and chemical change, 10 new “airs,” and observations that led to the discovery of photosynthesis. Due to his support for the principles of the French Revolution, Priestley’s house, library, and laboratory were destroyed by a mob in 1791. He and his wife emigrated to the United States in 1794 to join their three sons, who had previously emigrated to Pennsylvania. Priestley never returned to England and died in his new home in Pennsylvania.
Despite the pioneering studies of Priestley and others, a clear understanding of combustion remained elusive. In the late 18th century, however, the French scientist Antoine Lavoisier (1743–94) showed that combustion is the reaction of a carbon-containing substance with oxygen to form carbon dioxide and water and that life depends on a similar reaction, which today we call respiration. Lavoisier also wrote the first modern chemistry text and is widely regarded as the father of modern chemistry. His most important contribution was the law of conservation of mass33, which states that in any chemical reaction, the mass of the substances that react equals the mass of the products that are formed. That is, in a chemical reaction, mass is neither lost nor destroyed. Unfortunately, Lavoisier invested in a private corporation that collected taxes for the Crown, and royal tax collectors were not popular during the French Revolution. He was executed on the guillotine at age 51, prematurely terminating his contributions to chemistry.
The Atomic Theory of Matter
33. In any chemical reaction, the mass of the substances that react equals the mass of the products that are formed.
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In 1803, the English schoolteacher John Dalton (1766–1844) expanded Proust’s development of the law of definite proportions (Section 1.2 "The Scientific Method") and Lavoisier’s findings on the conservation of mass in chemical reactions to propose that elements consist of indivisible particles that he called atoms (taking the term from Democritus and Leucippus). Dalton’s atomic theory of matter contains four fundamental hypotheses:
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1. All matter is composed of tiny indivisible particles called atoms. 2. All atoms of an element are identical in mass and chemical properties, whereas atoms of different elements differ in mass and fundamental chemical properties. 3. A chemical compound is a substance that always contains the same atoms in the same ratio. 4. In chemical reactions, atoms from one or more compounds or elements redistribute or rearrange in relation to other atoms to form one or more new compounds. Atoms themselves do not undergo a change of identity in chemical reactions. This last hypothesis suggested that the alchemists’ goal of transmuting other elements to gold was impossible, at least through chemical reactions. We now know that Dalton’s atomic theory is essentially correct, with four minor modifications: 1. Not all atoms of an element must have precisely the same mass. 2. Atoms of one element can be transformed into another through nuclear reactions. 3. The compositions of many solid compounds are somewhat variable. 4. Under certain circumstances, some atoms can be divided (split into smaller particles). These modifications illustrate the effectiveness of the scientific method; later experiments and observations were used to refine Dalton’s original theory.
The Law of Multiple Proportions Despite the clarity of his thinking, Dalton could not use his theory to determine the elemental compositions of chemical compounds because he had no reliable scale of atomic masses; that is, he did not know the relative masses of elements such as carbon and oxygen. For example, he knew that the gas we now call carbon monoxide contained carbon and oxygen in the ratio 1:1.33 by mass, and a second compound, the gas we call carbon dioxide, contained carbon and oxygen in the ratio 1:2.66 by mass. Because 2.66/1.33 = 2.00, the second compound contained twice as many oxygen atoms per carbon atom as did the first. But what was the correct formula for each compound? If the first compound consisted of particles that contain one carbon atom and one oxygen atom, the second must consist of particles that contain one carbon atom and two oxygen atoms. If the first compound had two carbon atoms and one oxygen atom, the second must have two carbon atoms and two oxygen atoms. If the first had one carbon atom and two oxygen atoms, the second would have one carbon atom and four oxygen atoms, and so forth. Dalton had no way to distinguish among these or more complicated alternatives. However,
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these data led to a general statement that is now known as the law of multiple proportions34: when two elements form a series of compounds, the ratios of the masses of the second element that are present per gram of the first element can almost always be expressed as the ratios of integers. (The same law holds for mass ratios of compounds forming a series that contains more than two elements.) Example 4 shows how the law of multiple proportions can be applied to determine the identity of a compound.
34. When two elements form a series of compounds, the ratios of the masses of the second element that are present per gram of the first element can almost always be expressed as the ratios of integers. (The same law holds for the mass ratios of compounds forming a series that contains more than two elements.)
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EXAMPLE 4 A chemist is studying a series of simple compounds of carbon and hydrogen. The following table lists the masses of hydrogen that combine with 1 g of carbon to form each compound. Compound Mass of Hydrogen (g) A
0.0839
B
0.1678
C
0.2520
D
a. Determine whether these data follow the law of multiple proportions. b. Calculate the mass of hydrogen that would combine with 1 g of carbon to form D, the fourth compound in the series. Given: mass of hydrogen per gram of carbon for three compounds Asked for: a. ratios of masses of hydrogen to carbon b. mass of hydrogen per gram of carbon for fourth compound in series Strategy: A Select the lowest mass to use as the denominator and then calculate the ratio of each of the other masses to that mass. Include other ratios if appropriate. B If the ratios are small whole integers, the data follow the law of multiple proportions. C Decide whether the ratios form a numerical series. If so, then determine the next member of that series and predict the ratio corresponding to the next compound in the series. D Use proportions to calculate the mass of hydrogen per gram of carbon in that compound.
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Solution: A Compound A has the lowest mass of hydrogen, so we use it as the denominator. The ratios of the remaining masses of hydrogen, B and C, that combine with 1 g of carbon are as follows:
0.2520 g C 3 = = 3.00 = A 0.0839 g 1 0.1678 g B 2 = = 2.00 = A 0.0839 g 1 0.2520 g C 3 = = 1.502 ≈ B 0.1678 g 2 B The ratios of the masses of hydrogen that combine with 1 g of carbon are indeed composed of small whole integers (3/1, 2/1, 3/2), as predicted by the law of multiple proportions. C The ratios B/A and C/A form the series 2/1, 3/1, so the next member of the series should be D/A = 4/1. D Thus, if compound D exists, it would be formed by combining 4 × 0.0839 g = 0.336 g of hydrogen with 1 g of carbon. Such a compound does exist; it is methane, the major constituent of natural gas. Exercise Four compounds containing only sulfur and fluorine are known. The following table lists the masses of fluorine that combine with 1 g of sulfur to form each compound. Compound Mass of Fluorine (g)
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A
3.54
B
2.96
C
2.36
D
0.59
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a. Determine the ratios of the masses of fluorine that combine with 1 g of sulfur in these compounds. Are these data consistent with the law of multiple proportions? b. Calculate the mass of fluorine that would combine with 1 g of sulfur to form the next two compounds in the series: E and F. Answer: a. A/D = 6.0 or 6/1; B/D ≈ 5.0, or 5/1; C/D = 4.0, or 4/1; yes b. Ratios of 3.0 and 2.0 give 1.8 g and 1.2 g of fluorine/gram of sulfur, respectively. (Neither of these compounds is yet known.)
Avogadro’s Hypothesis In a further attempt to establish the formulas of chemical compounds, the French chemist Joseph Gay-Lussac (1778–1850) carried out a series of experiments using volume measurements. Under conditions of constant temperature and pressure, he carefully measured the volumes of gases that reacted to make a given chemical compound, together with the volumes of the products if they were gases. GayLussac found, for example, that one volume of chlorine gas always reacted with one volume of hydrogen gas to produce two volumes of hydrogen chloride gas. Similarly, one volume of oxygen gas always reacted with two volumes of hydrogen gas to produce two volumes of water vapor (part (a) in Figure 1.14 "Gay-Lussac’s Experiments with Chlorine Gas and Hydrogen Gas"). Figure 1.14 Gay-Lussac’s Experiments with Chlorine Gas and Hydrogen Gas
(a) One volume of chlorine gas reacted with one volume of hydrogen gas to produce two volumes of hydrogen chloride gas, and one volume of oxygen gas reacted with two volumes of hydrogen gas to produce two volumes of water vapor. (b) A summary of Avogadro’s hypothesis, which interpreted Gay-Lussac’s results in terms of atoms. Note that the simplest way for two molecules of hydrogen chloride to be produced is if hydrogen and chlorine each consist of molecules that contain two atoms of the element.
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Gay-Lussac’s results did not by themselves reveal the formulas for hydrogen chloride and water. The Italian chemist Amadeo Avogadro (1776–1856) developed the key insight that led to the exact formulas. He proposed that when gases are measured at the same temperature and pressure, equal volumes of different gases contain equal numbers of gas particles. Avogadro’s hypothesis, which explained GayLussac’s results, is summarized here and in part (b) in Figure 1.14 "Gay-Lussac’s Experiments with Chlorine Gas and Hydrogen Gas":
one volume two volumes one volume (or particle) of + (or particle) of → (or particles) of hydrogen hydrogen chloride chlorine If Dalton’s theory of atoms was correct, then each particle of hydrogen or chlorine had to contain at least two atoms of hydrogen or chlorine because two particles of hydrogen chloride were produced. The simplest—but not the only—explanation was that hydrogen and chlorine contained two atoms each (i.e., they were diatomic) and that hydrogen chloride contained one atom each of hydrogen and chlorine. Applying this reasoning to Gay-Lussac’s results with hydrogen and oxygen leads to the conclusion that water contains two hydrogen atoms per oxygen atom. Unfortunately, because no data supported Avogadro’s hypothesis that equal volumes of gases contained equal numbers of particles, his explanations and formulas for simple compounds were not generally accepted for more than 50 years. Dalton and many others continued to believe that water particles contained one hydrogen atom and one oxygen atom, rather than two hydrogen atoms and one oxygen atom. The historical development of the concept of the atom is summarized in Figure 1.15 "A Summary of the Historical Development of the Concept of the Atom".
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Figure 1.15 A Summary of the Historical Development of the Concept of the Atom
Summary The ancient Greeks first proposed that matter consisted of fundamental particles called atoms. Chemistry took its present scientific form in the 18th century, when careful quantitative experiments by Lavoisier, Proust, and Dalton resulted in the law of definite proportions, the law of conservation of mass, and the law of multiple proportions, which laid the groundwork for Dalton’s atomic theory of matter. In particular, Avogadro’s hypothesis provided the first link between the macroscopic properties of a substance (in this case, the volume of a gas) and the number of atoms or molecules present.
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KEY TAKEAWAY • The development of the atomic model relied on the application of the scientific method over several centuries.
CONCEPTUAL PROBLEMS 1. Define combustion and discuss the contributions made by Priestley and Lavoisier toward understanding a combustion reaction. 2. Chemical engineers frequently use the concept of “mass balance” in their calculations, in which the mass of the reactants must equal the mass of the products. What law supports this practice? 3. Does the law of multiple proportions apply to both mass ratios and atomic ratios? Why or why not? 4. What are the four hypotheses of the atomic theory of matter? 5. Much of the energy in France is provided by nuclear reactions. Are such reactions consistent with Dalton’s hypotheses? Why or why not? 6. Does 1 L of air contain the same number of particles as 1 L of nitrogen gas? Explain your answer.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Numerical Problems. 1. One of the minerals found in soil has an Al:Si:O atomic ratio of 0.2:0.2:0.5. Is this consistent with the law of multiple proportions? Why or why not? Is the ratio of elements consistent with Dalton’s atomic theory of matter? 2. Nitrogen and oxygen react to form three different compounds that contain 0.571 g, 1.143 g, and 2.285 g of oxygen/gram of nitrogen, respectively. Is this consistent with the law of multiple proportions? Explain your answer. 3. Three binary compounds of vanadium and oxygen are known. The following table gives the masses of oxygen that combine with 10.00 g of vanadium to form each compound. Compound Mass of Oxygen (g) A
4.71
B
6.27
C a. Determine the ratio of the masses of oxygen that combine with 3.14 g of vanadium in compounds A and B. b. Predict the mass of oxygen that would combine with 3.14 g of vanadium to form the third compound in the series. 4. Three compounds containing titanium, magnesium, and oxygen are known. The following table gives the masses of titanium and magnesium that react with 5.00 g of oxygen to form each compound. Compound Mass of Titanium (g) Mass of Magnesium (g) A
4.99
2.53
B
3.74
3.80
C a. Determine the ratios of the masses of titanium and magnesium that combine with 5.00 g of oxygen in these compounds. b. Predict the masses of titanium and magnesium that would combine with 5.00 g of oxygen to form another possible compound in the series: C.
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1.5 The Atom LEARNING OBJECTIVE 1. To become familiar with the components and structure of the atom.
To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons35, protons36, and neutrons37.This is an oversimplification that ignores the other subatomic particles that have been discovered, but it is sufficient for our discussion of chemical principles. Some properties of these subatomic particles are summarized in Table 1.3 "Properties of Subatomic Particles*", which illustrates three important points. 1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. We usually assign relative charges of −1 and +1 to the electron and proton, respectively. 2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral. 3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute by far the bulk of the mass of atoms.
35. A subatomic particle with a negative charge that resides around the nucleus of all atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
36. A subatomic particle with a positive charge that resides in the nucleus of all atoms. 37. A subatomic particle with no charge that resides in the nucleus of almost all atoms.
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Table 1.3 Properties of Subatomic Particles* Atomic Mass (amu)
Electrical Charge (coulombs)
electron 9.109 × 10−28
0.0005486
−1.602 × 10−19
−1
1.673 × 10−24
1.007276
+1.602 × 10−19
+1
neutron 1.675 × 10−24
1.008665
0
0
Particle
proton
Mass (g)
Relative Charge
* For a review of using scientific notation and units of measurement, see Essential Skills 1 (Section 1.9 "Essential Skills 1").
The Electron Long before the end of the 19th century, it was well Figure 1.16 A Gas Discharge known that applying a high voltage to a gas contained at Tube Producing Cathode low pressure in a sealed tube (called a gas discharge Rays tube) caused electricity to flow through the gas, which then emitted light (Figure 1.16 "A Gas Discharge Tube Producing Cathode Rays"). Researchers trying to understand this phenomenon found that an unusual form of energy was also emitted from the cathode, or negatively charged electrode; hence this form of energy was called cathode rays. In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the ultimate form of matter. He demonstrated that cathode When a high voltage is applied to a gas contained at low pressure rays could be deflected, or bent, by magnetic or electric in a gas discharge tube, fields, which indicated that cathode rays consist of electricity flows through the gas, and energy is emitted in the form charged particles (Figure 1.17 "Deflection of Cathode of light. Rays by an Electric Field"). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; we now call these particles electrons. Most important for chemistry, Thomson found that the mass-to-charge ratio of cathode rays was independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
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Figure 1.17 Deflection of Cathode Rays by an Electric Field
As the cathode rays travel toward the right, they are deflected toward the positive electrode (+), demonstrating that they are negatively charged.
Subsequently, the American scientist Robert Millikan (1868–1953) carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron. With this information and Thomson’s mass-to-charge ratio, Millikan determined the mass of an electron:
mass charge
× charge = mass
It was at this point that two separate lines of investigation began to converge, both aimed at determining how and why matter emits energy.
Radioactivity
38. The spontaneous emission of energy rays (radiation) by matter.
1.5 The Atom
The second line of investigation began in 1896, when the French physicist Henri Becquerel (1852–1908) discovered that certain minerals, such as uranium salts, emitted a new form of energy. Becquerel’s work was greatly extended by Marie Curie (1867–1934) and her husband, Pierre (1854–1906); all three shared the Nobel Prize in Physics in 1903. Marie Curie coined the term radioactivity38 (from the Latin radius, meaning “ray”) to describe the emission of energy rays by matter. She found that one particular uranium ore, pitchblende, was substantially more radioactive than most, which suggested that it contained one or more highly radioactive impurities. Starting with several tons of pitchblende, the Curies isolated two new radioactive elements after months of work: polonium, which was named for Marie’s native Poland, and radium, which was named for its intense radioactivity. Pierre Curie carried a vial of radium in his coat pocket to demonstrate its greenish glow, a habit that caused him to become ill from radiation poisoning
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well before he was run over by a horse-drawn wagon and killed instantly in 1906. Marie Curie, in turn, died of what was almost certainly radiation poisoning. Building on the Curies’ work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a Radium bromide illuminated by its own radioactive glow. positive charge and were massive compared to electrons. Because it was the first kind of radiation to be This 1922 photo was taken in the dark in the Curie laboratory. discovered, Rutherford called these substances α particles. Rutherford also showed that the particles in the second type of radiation, β particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. A third type of radiation, γ rays, was discovered somewhat later and found to be similar to a lower-energy form of radiation called x-rays, now used to produce images of bones and teeth. These three kinds of radiation—α particles, β particles, and γ rays—are readily distinguished by the way they are deflected by an electric field and by the degree to which they penetrate matter. As Figure 1.18 "Effect of an Electric Field on α Particles, β Particles, and γ Rays" illustrates, α particles and β particles are deflected in opposite directions; α particles are deflected to a much lesser extent because of their higher mass-to-charge ratio. In contrast, γ rays have no charge, so they are not deflected by electric or magnetic fields. Figure 1.19 "Relative Penetrating Power of the Three Types of Radiation" shows that α particles have the least penetrating power and are stopped by a sheet of paper, whereas β particles can pass through thin sheets of metal but are absorbed by lead foil or even thick glass. In contrast, γ-rays can readily penetrate matter; thick blocks of lead or concrete are needed to stop them.
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Figure 1.18 Effect of an Electric Field on α Particles, β Particles, and γ Rays
A negative electrode deflects negatively charged β particles, whereas a positive electrode deflects positively charged α particles. Uncharged γ rays are unaffected by an electric field. (Relative deflections are not shown to scale.)
Figure 1.19 Relative Penetrating Power of the Three Types of Radiation
A sheet of paper stops comparatively massive α particles, whereas β particles easily penetrate paper but are stopped by a thin piece of lead foil. Uncharged γ rays penetrate the paper and lead foil; a much thicker piece of lead or concrete is needed to absorb them.
The Atomic Model Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of
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the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure 1.20 "Thomson’s Plum Pudding or Chocolate Chip Cookie Model of the Atom"). In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the Figure 1.20 Thomson’s Plum atom was impossible. Rutherford aimed a stream of α Pudding or Chocolate Chip particles at a very thin gold foil target (part (a) in Figure Cookie Model of the Atom 1.21 "A Summary of Rutherford’s Experiments") and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets with a thickness that minimized the number of atoms in the target. If Thomson’s model of the atom were correct, the positively charged α particles should crash through the uniformly distributed mass of the gold target like In this model, the electrons are cannonballs through the side of a wooden house. They embedded in a uniform sphere of might be moving a little slower when they emerged, but positive charge. they should pass essentially straight through the target (part (b) in Figure 1.21 "A Summary of Rutherford’s Experiments"). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (part (c) in Figure 1.21 "A Summary of Rutherford’s Experiments"). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Figure 1.21 A Summary of Rutherford’s Experiments
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(a) A representation of the apparatus Rutherford used to detect deflections in a stream of α particles aimed at a thin gold foil target. The particles were produced by a sample of radium. (b) If Thomson’s model of the atom were correct, the α particles should have passed straight through the gold foil. (c) But a small number of α particles were deflected in various directions, including right back at the source. This could be true only if the positive charge were much more massive than the α particle. It suggested that the mass of the gold atom is concentrated in a very small region of space, which he called the nucleus.
Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus39. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source. Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today we know that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he thought that physics was superior to chemistry. In his opinion, “All science is either physics or stamp collecting.” (The authors of this text do not share Rutherford’s view!) Subsequently, Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons and is therefore simply the nucleus of a helium atom.
39. The central core of an atom where protons and any neutrons reside.
1.5 The Atom
The historical development of the different models of the atom’s structure is summarized in Figure 1.22 "A Summary of the Historical Development of Models of the Components and Structure of the Atom". Rutherford’s model of the atom is essentially the same as the modern one, except that we now know that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are
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distributed according to a set of principles described in Chapter 6 "The Structure of Atoms". Figure 1.23 "The Evolution of Atomic Theory, as Illustrated by Models of the Oxygen Atom" shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today. Figure 1.22 A Summary of the Historical Development of Models of the Components and Structure of the Atom
The dates in parentheses are the years in which the key experiments were performed.
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Figure 1.23 The Evolution of Atomic Theory, as Illustrated by Models of the Oxygen Atom
Bohr’s model and the current model are described in Chapter 6 "The Structure of Atoms".
Summary Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy).
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KEY TAKEAWAY • The atom consists of discrete particles that govern its chemical and physical behavior.
CONCEPTUAL PROBLEMS 1. Describe the experiment that provided evidence that the proton is positively charged. 2. What observation led Rutherford to propose the existence of the neutron? 3. What is the difference between Rutherford’s model of the atom and the model chemists use today? 4. If cathode rays are not deflected when they pass through a region of space, what does this imply about the presence or absence of a magnetic field perpendicular to the path of the rays in that region? 5. Describe the outcome that would be expected from Rutherford’s experiment if the charge on α particles had remained the same but the nucleus were negatively charged. If the nucleus were neutral, what would have been the outcome? 6. Describe the differences between an α particle, a β particle, and a γ ray. Which has the greatest ability to penetrate matter?
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Numerical Problems. 1. Using the data in Table 1.3 "Properties of Subatomic Particles*" and the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"), calculate the percentage of the mass of a silicon atom that is due to a. electrons. b. protons. 2. Using the data in Table 1.3 "Properties of Subatomic Particles*" and the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"), calculate the percentage of the mass of a helium atom that is due to a. electrons. b. protons. 3. The radius of an atom is approximately 104 times larger than the radius of its nucleus. If the radius of the nucleus were 1.0 cm, what would be the radius of the atom in centimeters? in miles? 4. The total charge on an oil drop was found to be 3.84 × 10−18 coulombs. What is the total number of electrons contained in the drop?
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1.6 Isotopes and Atomic Masses LEARNING OBJECTIVE 1. To know the meaning of isotopes and atomic masses.
Rutherford’s nuclear model of the atom helped explain why atoms of different elements exhibit different chemical behavior. The identity of an element is defined by its atomic number (Z)40, the number of protons in the nucleus of an atom of the element. The atomic number is therefore different for each element. The known elements are arranged in order of increasing Z in the periodic table41 (Figure 1.24 "The Periodic Table Showing the Elements in Order of Increasing "; also see Chapter 32 "Appendix H: Periodic Table of Elements"),We will explain the rationale for the peculiar format of the periodic table in Chapter 7 "The Periodic Table and Periodic Trends". in which each element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons. In a neutral atom, the number of electrons equals the number of protons. Figure 1.24 The Periodic Table Showing the Elements in Order of Increasing Z
40. The number of protons in the nucleus of an atom of an element. 41. A chart of the chemical elements arranged in rows of increasing atomic number so that the elements in each column (group) have similar chemical properties.
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As described in Section 1.7 "Introduction to the Periodic Table", the metals are on the bottom left in the periodic table, and the nonmetals are at the top right. The semimetals lie along a diagonal line separating the metals and nonmetals.
In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to remind us of their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table 1.4 "Element Symbols Based on Names No Longer in Use". As you work through this text, you will encounter the names and symbols of the elements repeatedly, and much as you become familiar with characters in a play or a film, their names and symbols will become familiar. Table 1.4 Element Symbols Based on Names No Longer in Use Element
1.6 Isotopes and Atomic Masses
Symbol
Derivation
Meaning
antimony
Sb
stibium
Latin for “mark”
copper
Cu
cuprum
from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold
Au
aurum
Latin for “gold”
iron
Fe
ferrum
Latin for “iron”
lead
Pb
plumbum
Latin for “heavy”
mercury
Hg
hydrargyrum Latin for “liquid silver”
potassium K
kalium
from the Arabic al-qili, “alkali”
silver
Ag
argentum
Latin for “silver”
sodium
Na
natrium
Latin for “sodium”
tin
Sn
stannum
Latin for “tin”
tungsten
W
wolfram
German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin
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Recall from Section 1.5 "The Atom" that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes42. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A)43, the sum of the numbers of protons and neutrons. The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carboncontaining material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as A Z X, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore 12 6 C. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, 12 12 6 C is more often written as C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z, as described in Chapter 20 "Nuclear Chemistry".
42. Atoms that have the same numbers of protons but different numbers of neutrons. 43. The number of protons and neutrons in the nucleus of an atom of an element.
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13 In addition to 12C, a typical sample of carbon contains 1.11% 13 6 C ( C), with 7 14 neutrons and 6 protons, and a trace of 14 6 C ( C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archaeology (see Chapter 14 "Chemical Kinetics"). Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table 1.5 "Properties of Selected Isotopes".
Table 1.5 Properties of Selected Isotopes
Element
Symbol
hydrogen H
Atomic Mass (amu) 1.0079
boron
B
10.81
carbon
C
12.011
Isotope Mass Number 1 2
O
10 11
Fe
12 13
U
12 (defined) 13.003355 15.994915
Percent Abundances (%) 99.9855 0.0115 19.91 80.09 99.89 1.11 99.757
15.9994 17
16.999132
0.0378
18
17.999161
0.205
55.845
53.939611
5.82
56
55.934938
91.66
57
56.935394
2.19
58
57.933276
0.33
234 uranium
10.012937 11.009305
54 iron
1.007825 2.014102
16 oxygen
Isotope Masses (amu)
234.040952
0.0054
238.03 235
235.043930
0.7204
238
238.050788
99.274
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
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EXAMPLE 5 An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes. Given: number of protons and neutrons Asked for: element and atomic symbol Strategy: A Refer to the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and use the number of protons to identify the element. B Calculate the mass number of each isotope by adding together the numbers of protons and neutrons. C Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element. Solution: A The element with 82 protons (atomic number of 82) is lead: Pb. B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, 206
207
respectively. The symbols for these isotopes are 82 P b, 82 which are usually abbreviated as 206Pb, 207Pb, and 208Pb.
P b, and 208 82 P b,
Exercise Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons. 79
81
Answer: 35 B r and 35 B r or, more commonly, 79Br and 81Br.
Although the masses of the electron, the proton, and the neutron are known to a high degree of precision (Table 1.3 "Properties of Subatomic Particles*"), the mass
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of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions (Chapter 20 "Nuclear Chemistry"). Because atoms are much too small to measure individually and do not have a charge, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions44. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure 1.25 "Determining Relative Atomic Masses Using a Mass Spectrometer"). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest. Figure 1.25 Determining Relative Atomic Masses Using a Mass Spectrometer
44. A charged particle produced when one or more electrons is removed from or added to an atom or molecule.
1.6 Isotopes and Atomic Masses
Chlorine consists of two isotopes, 35Cl and 37Cl, in approximately a 3:1 ratio. (a) When a sample of elemental chlorine is injected into the mass spectrometer, electrical energy is used to dissociate the Cl 2 molecules into chlorine atoms and convert the chlorine atoms to Cl+ ions. The ions are then accelerated into a magnetic field. The extent to which
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the ions are deflected by the magnetic field depends on their relative mass-to-charge ratios. Note that the lighter 35 + Cl ions are deflected more than the heavier 37Cl+ ions. By measuring the relative deflections of the ions, chemists can determine their mass-to-charge ratios and thus their masses. (b) Each peak in the mass spectrum corresponds to an ion with a particular mass-to-charge ratio. The abundance of the two isotopes can be determined from the heights of the peaks.
The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu)45, defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12 C is the only atom listed in Table 1.5 "Properties of Selected Isotopes" whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g. Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2 H to the mass of 12C, so the absolute mass of 2H is
mass of 2 H mass of 12 C
× mass of 12 C = 0.167842 × 12 amu = 2.104104 amu
The masses of the other elements are determined in a similar way. The periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") lists the atomic masses of all the elements. If you compare these values with those given for some of the isotopes in Table 1.5 "Properties of Selected Isotopes", you can see that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes. Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as (0.9889 × 12 amu) + (0.0111 × 13.003355 amu) = 12.01 amu Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with our calculation. The value of 12.01 is shown under the symbol for C in the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"), although without the atom of C; −24 1 amu = 1. 66 × 10 g. abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of
45. One-twelfth of the mass of one 12
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carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.
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EXAMPLE 6 Naturally occurring bromine consists of the two isotopes listed in the following table: Isotope Exact Mass (amu) Percent Abundance (%) 79
Br
78.9183
50.69
81
Br
80.9163
49.31
Calculate the atomic mass of bromine. Given: exact mass and percent abundance Asked for: atomic mass Strategy: A Convert the percent abundances to decimal form to obtain the mass fraction of each isotope. B Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass. C Add together the weighted masses to obtain the atomic mass of the element. D Check to make sure that your answer makes sense. Solution: A The atomic mass is the weighted average of the masses of the isotopes. In general, we can write atomic mass of element = [(mass of isotope 1 in amu) (mass fraction of isotope 1)] + [(mass of isotope 2) (mass fraction of isotope 2)] + … Bromine has only two isotopes. Converting the percent abundances to mass fractions gives
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50.69 = 0.5069 100 49.31 81 Br: = 0.4931 100 79
Br:
B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass: 79
Br: 79.9183 amu × 0.5069 = 40.00 amu
81
Br: 80.9163 amu × 0.4931 = 39.90 amu C The sum of the weighted masses is the atomic mass of bromine is
40.00 amu + 39.90 amu = 79.90 amu D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%. Exercise Magnesium has the three isotopes listed in the following table: Isotope Exact Mass (amu) Percent Abundance (%) 24
Mg
23.98504
78.70
25
Mg
24.98584
10.13
26
Mg
25.98259
11.17
Use these data to calculate the atomic mass of magnesium. Answer: 24.31 amu
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Summary Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol.
KEY TAKEAWAY • The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge.
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CONCEPTUAL PROBLEMS 1. Complete the following table for the missing elements, symbols, and numbers of electrons. Element
Symbol Number of Electrons
molybdenum 19 titanium B 53 Sm helium 14 2. Complete the following table for the missing elements, symbols, and numbers of electrons. Element
Symbol Number of Electrons
lanthanum Ir aluminum 80 sodium Si 9 Be 3. Is the mass of an ion the same as the mass of its parent atom? Explain your answer. 4. What isotopic standard is used for determining the mass of an atom? A
5. Give the symbol Z X for these elements, all of which exist as a single isotope. a. beryllium
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b. c. d. e. f. g. h. i.
ruthenium phosphorus aluminum cesium praseodymium cobalt yttrium arsenic A
6. Give the symbol Z X for these elements, all of which exist as a single isotope. a. b. c. d. e. f. g. h. i.
fluorine helium terbium iodine gold scandium sodium niobium manganese
7. Identify each element, represented by X, that have the given symbols. 55
a. 26 X 74 b. 33 X 24
c. 12 X 127 d. 53 X 40
e. 18 X 152 f. 63 X
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Numerical Problems. 1. The isotopes 131I and 60Co are commonly used in medicine. Determine the number of neutrons, protons, and electrons in a neutral atom of each. 2. Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope: a. b. c. d.
97Tc 113In 63Ni 55Fe
3. Both technetium-97 and americium-240 are produced in nuclear reactors. Determine the number of protons, neutrons, and electrons in the neutral atoms of each. 4. The following isotopes are important in archaeological research. How many protons, neutrons, and electrons does a neutral atom of each contain? a. b. c. d. e.
207Pb 16O 40K 137Cs 40Ar
5. Copper, an excellent conductor of heat, has two isotopes: 63Cu and 65Cu. Use the following information to calculate the average atomic mass of copper: Isotope Percent Abundance (%) Atomic Mass (amu) 63Cu
69.09
62.9298
65Cu
30.92
64.9278
6. Silicon consists of three isotopes with the following percent abundances: Isotope Percent Abundance (%) Atomic Mass (amu) 28Si
92.18
27.976926
29Si
4.71
28.976495
30Si
3.12
29.973770
Calculate the average atomic mass of silicon.
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7. Complete the following table for neon. The average atomic mass of neon is 20.1797 amu. Isotope Percent Abundance (%) Atomic Mass (amu) 20Ne
90.92
19.99244
21Ne
0.257
20.99395
22Ne 63
62
8. Are 28 X and 29 X isotopes of the same element? Explain your answer. 9. Complete the following table: Isotope
Number of Protons
Number of Neutrons
Number of Electrons
238X
95
238U 75
112
10. Complete the following table: Isotope
Number of Protons
Number of Neutrons
Number of Electrons
57Fe 40X
20
36S 11. Using a mass spectrometer, a scientist determined the percent abundances of the isotopes of sulfur to be 95.27% for 32S, 0.51% for 33S, and 4.22% for 34S. Use the atomic mass of sulfur from the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and the following atomic masses to determine whether these data are accurate, assuming that these are the only isotopes of sulfur: 31.972071 amu for 32S, 32.971459 amu for 33S, and 33.967867 amu for 34S. 12. The percent abundances of two of the three isotopes of oxygen are 99.76% for 16O, and 0.204% for 18O. Use the atomic mass of oxygen given in the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements") and the following data to determine the mass of 17O: 15.994915 amu for 16O and 17.999160 amu for 18O. 13. Which element has the higher proportion by mass in NaI?
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14. Which element has the higher proportion by mass in KBr?
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1.7 Introduction to the Periodic Table 46. A row of elements in the periodic table. 47. A vertical column of elements in the periodic table. Elements with similar chemical properties reside in the same group. 48. Any element in groups 1, 2, and 13–18 in the periodic table. These groups contain metals, semimetals, and nonmetals. 49. Any element in groups 3–12 in the periodic table. All of the transition elements are metals. 50. Any element to the left of the zigzag line in the periodic table that runs from boron to astatine. All metals except mercury are solids at room temperature and pressure. 51. Any element to the right of the zigzag line in the periodic table that runs from boron to astatine. Nonmetals may be solids, liquids, or gases at room temperature and pressure. 52. Any element that lies adjacent to the zigzag line in the periodic table that runs from boron to astatine. Semimetals (also called metalloids) exhibit properties intermediate between those of metals and nonmetals. 53. The ability to be pulled into wires. Metals are ductile, whereas nonmetals are usually brittle. 54. The ability to be hammered or pressed into thin sheets or foils. Metals are malleable, whereas nonmetals are usually brittle. 55. Having a shiny appearance. Metals are lustrous, whereas nonmetals are not.
LEARNING OBJECTIVE 1. To become familiar with the organization of the periodic table.
The elements are arranged in a periodic table (Figure 1.24 "The Periodic Table Showing the Elements in Order of Increasing "; also see Chapter 32 "Appendix H: Periodic Table of Elements"), which is probably the single most important learning aid in chemistry. It summarizes huge amounts of information about the elements in a way that permits you to predict many of their properties and chemical reactions. The elements are arranged in seven horizontal rows, in order of increasing atomic number from left to right and top to bottom. The rows are called periods46, and they are numbered from 1 to 7. The elements are stacked in such a way that elements with similar chemical properties form vertical columns, called groups47, numbered from 1 to 18 (older periodic tables use a system based on roman numerals). Groups 1, 2, and 13–18 are the main group elements48, listed as A in older tables. Groups 3–12 are in the middle of the periodic table and are the transition elements49, listed as B in older tables. The two rows of 14 elements at the bottom of the periodic table are the lanthanides and the actinides, whose positions in the periodic table are indicated in group 3. A more comprehensive description of the periodic table is found in Chapter 7 "The Periodic Table and Periodic Trends".
Metals, Nonmetals, and Semimetals The heavy orange zigzag line running diagonally from the upper left to the lower right through groups 13–16 in Figure 1.24 "The Periodic Table Showing the Elements in Order of Increasing " divides the elements into metals50 (in blue, below and to the left of the line) and nonmetals51 (in bronze, above and to the right of the line). As you might expect, elements colored in gold that lie along the diagonal line exhibit properties intermediate between metals and nonmetals; they are called semimetals52. The distinction between metals and nonmetals is one of the most fundamental in chemistry. Metals—such as copper or gold—are good conductors of electricity and heat; they can be pulled into wires because they are ductile53; they can be hammered or pressed into thin sheets or foils because they are malleable54; and most have a shiny appearance, so they are lustrous55. The vast majority of the
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known elements are metals. Of the metals, only mercury is a liquid at room temperature and pressure; all the rest are solids. Nonmetals, in contrast, are generally poor conductors of heat and electricity and are not lustrous. Nonmetals can be gases (such as chlorine), liquids (such as bromine), or solids (such as iodine) at room temperature and pressure. Most solid nonmetals are brittle, so they break into small pieces when hit with a hammer or pulled into a wire. As expected, semimetals exhibit properties intermediate between metals and nonmetals.
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EXAMPLE 7 Based on its position in the periodic table, do you expect selenium to be a metal, a nonmetal, or a semimetal? Given: element Asked for: classification Strategy: Find selenium in the periodic table shown in Figure 1.24 "The Periodic Table Showing the Elements in Order of Increasing " and then classify the element according to its location. Solution: The atomic number of selenium is 34, which places it in period 4 and group 16. In Figure 1.24 "The Periodic Table Showing the Elements in Order of Increasing ", selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metalnonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties. Exercise Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a semimetal? Answer: metal 56. Any element in group 1 of the periodic table. 57. Any element in group 2 of the periodic table.
Descriptive Names
58. Derived from the Greek for “salt forming,” an element in group 17 of the periodic table.
As we noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins. For example, the elements of group 1 are known as the alkali metals56, group 2 are the alkaline earth metals57, group 17 are the halogens58, and group 18 are the noble gases59.
59. Any element in group 18 of the periodic table. All are unreactive monatomic gases at room temperature and pressure.
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The Alkali Metals The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in group 1, but it is not a metal. The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.
The Alkaline Earth Metals The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rather rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.
The Halogens The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt). Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.
The Noble Gases
60. A species containing a single atom.
The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are monatomic60. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the
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first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.
Note the Pattern The noble gases are unreactive at room temperature and pressure.
Summary The periodic table is an arrangement of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely used common names, including the alkali metals (group 1) and the alkaline earth metals (group 2) on the far left, and the halogens (group 17) and the noble gases (group 18) on the far right. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semimetals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable), whereas solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, the lanthanides, and the actinides.
KEY TAKEAWAY • The periodic table is used as a predictive tool.
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CONCEPTUAL PROBLEMS 1. Classify each element in Conceptual Problem 1 (Section 1.6 "Isotopes and Atomic Masses") as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. 2. Classify each element in Conceptual Problem 2 (Section 1.6 "Isotopes and Atomic Masses") as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. 3. Classify each element as a metal, a semimetal, or a nonmetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. a. b. c. d. e. f. g. h. i.
iron tantalum sulfur silicon chlorine nickel potassium radon zirconium
4. Which of these sets of elements are all in the same period? a. b. c. d.
potassium, vanadium, and ruthenium lithium, carbon, and chlorine sodium, magnesium, and sulfur chromium, nickel, and krypton
5. Which of these sets of elements are all in the same period? a. b. c. d.
barium, tungsten, and argon yttrium, zirconium, and selenium potassium, calcium, and zinc scandium, bromine, and manganese
6. Which of these sets of elements are all in the same group? a. b. c. d.
sodium, rubidium, and barium nitrogen, phosphorus, and bismuth copper, silver, and gold magnesium, strontium, and samarium
7. Which of these sets of elements are all in the same group? a. iron, ruthenium, and osmium
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b. nickel, palladium, and lead c. iodine, fluorine, and oxygen d. boron, aluminum, and gallium 8. Indicate whether each element is a transition metal, a halogen, or a noble gas. a. b. c. d. e. f. g. h. i. j. k. l.
manganese iridium fluorine xenon lithium carbon zinc sodium tantalum hafnium antimony cadmium
9. Which of the elements indicated in color in the periodic table shown below is most likely to exist as a monoatomic gas? As a diatomic gas? Which is most likely to be a semimetal? A reactive metal?
10. Based on their locations in the periodic table, would you expect these elements to be malleable? Why or why not? a. b. c. d. e. f. g.
phosphorus chromium rubidium copper aluminum bismuth neodymium
11. Based on their locations in the periodic table, would you expect these elements to be lustrous? Why or why not? a. b. c. d.
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sulfur vanadium nickel arsenic
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e. strontium f. cerium g. sodium
ANSWER 3.
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Symbol
Type
Fe
metal: transition metal
Ta
metal: transition metal
S
nonmetal
Si
semimetal
Cl
nonmetal (halogen)
Ni
metal: transition metal
K
metal: alkali metal
Rn
nonmetal (noble gas)
Zr
metal: transition metal
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1.8 Essential Elements for Life LEARNING OBJECTIVE 1. To understand the importance of elements to nutrition.
Of the approximately 115 elements known, only the 19 highlighted in purple in Figure 1.26 "The Essential Elements in the Periodic Table" are absolutely required in the human diet. These elements—called essential elements61—are restricted to the first four rows of the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"), with only two or three exceptions (molybdenum, iodine, and possibly tin in the fifth row). Some other elements are essential for specific organisms. For example, boron is required for the growth of certain plants, bromine is widely distributed in marine organisms, and tungsten is necessary for some microorganisms.
Figure 1.26 The Essential Elements in the Periodic Table
Elements that are known to be essential for human life are shown in purple; elements that are suggested to be essential are shown in green. Elements not known to be essential are shown in gray.
61. Any of the 19 elements that are absolutely required in the human diet for survival. An additional seven elements are thought to be essential for humans.
What makes an element “essential”? By definition, an essential element is one that is required for life and whose absence results in death. Because of the experimental difficulties involved in producing deficiencies severe enough to cause death,
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especially for elements that are required in very low concentrations in the diet, a somewhat broader definition is generally used. An element is considered to be essential if a deficiency consistently causes abnormal development or functioning and if dietary supplementation of that element—and only that element—prevents this adverse effect. Scientists determine whether an element is essential by raising rats, chicks, and other animals on a synthetic diet that has been carefully analyzed and supplemented with acceptable levels of all elements except the element of interest (E). Ultraclean environments, in which plastic cages are used and dust from the air is carefully removed, minimize inadvertent contamination. If the animals grow normally on a diet that is as low as possible in E, then either E is not an essential element or the diet is not yet below the minimum required concentration. If the animals do not grow normally on a low-E diet, then their diets are supplemented with E until a level is reached at which the animals grow normally. This level is the minimum required intake of element E.
Classification of the Essential Elements The approximate elemental composition of a healthy 70.0 kg (154 lb) adult human is listed in Table 1.6 "Approximate Elemental Composition of a Typical 70 kg Human". Note that most living matter consists primarily of the so-called bulk elements: oxygen, carbon, hydrogen, nitrogen, and sulfur—the building blocks of the compounds that constitute our organs and muscles. These five elements also constitute the bulk of our diet; tens of grams per day are required for humans. Six other elements—sodium, magnesium, potassium, calcium, chlorine, and phosphorus—are often referred to as macrominerals because they provide essential ions in body fluids and form the major structural components of the body. In addition, phosphorus is a key constituent of both DNA and RNA: the genetic building blocks of living organisms. The six macrominerals are present in the body in somewhat smaller amounts than the bulk elements, so correspondingly lower levels are required in the diet. The remaining essential elements—called trace elements—are present in very small amounts, ranging from a few grams to a few milligrams in an adult human. Finally, measurable levels of some elements are found in humans but are not required for growth or good health. Examples are rubidium and strontium, whose chemistry is similar to that of the elements immediately above them in the periodic table (potassium and calcium, respectively, which are essential elements). Because the body’s mechanisms for extracting potassium and calcium from foods are not 100% selective, small amounts of rubidium and strontium, which have no known biological function, are absorbed. Table 1.6 Approximate Elemental Composition of a Typical 70 kg Human Bulk Elements (kg) Macrominerals (g)
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oxygen carbon
44 calcium
1700
12.6 phosphorus
680
hydrogen
6.6 potassium
250
nitrogen
1.8 chlorine
115
sulfur
0.1 sodium
70
magnesium
42
Trace Elements (mg) iron
5000 lead
35
silicon
3000 barium
21
zinc
1750 molybdenum
14
rubidium
360 boron
14
copper
280 arsenic
~3
strontium
280 cobalt
~3
bromine
140 chromium
~3
tin
140 nickel
~3
manganese
70 selenium
~2
iodine
70 lithium
~2
aluminum
35 vanadium
~2
The Trace Elements Because it is difficult to detect low levels of some essential elements, the trace elements were relatively slow to be recognized as essential. Iron was the first. In the 17th century, anemia was proved to be caused by an iron deficiency and often was cured by supplementing the diet with extracts of rusty nails. It was not until the 19th century, however, that trace amounts of iodine were found to eliminate goiter (an enlarged thyroid gland). This is why common table salt is “iodized”: a small amount of iodine is added. Copper was shown to be essential for humans in 1928, and manganese, zinc, and cobalt soon after that. Molybdenum was not known to be an essential element until 1953, and the need for chromium, selenium, vanadium, fluorine, and silicon was demonstrated only in the last 50 years. It seems likely that in the future other elements, possibly including tin, will be found to be essential at very low levels.
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Many compounds of trace elements, such as arsenic, selenium, and chromium, are toxic and can even cause cancer, yet these elements are identified as essential elements in Figure 1.26 "The Essential Elements in the Periodic Table". In fact, there is some evidence that one bacterium has replaced phosphorus with arsenic, although the finding is controversial. This has opened up the possibility of a “shadow biosphere” on Earth in which life evolved from an as yet undetected common ancestor. How can elements toxic to life be essential? First, the toxicity of an element often depends on its chemical form—for example, only certain compounds of chromium are toxic, whereas others are used in mineral supplements. Second, as shown in Figure 1.27 "Possible Concentrations of an Essential Element in the Diet", every element has three possible levels of dietary intake: deficient, optimum, and toxic in order of increasing concentration in the diet. Very low intake levels lead to symptoms of deficiency. Over some range of higher intake levels, an organism is able to maintain its tissue concentrations of the element at a level that optimizes biological functions. Finally, at some higher intake level, the normal regulatory mechanisms are overloaded, causing toxic symptoms to appear. Each element has its own characteristic curve. Both the width of the plateau and the specific concentration corresponding to the center of the plateau region differ by as much as several orders of magnitude for different elements. In the adult human, for example, the recommended daily dietary intake is 10–18 mg of iron, 2–3 mg of copper, and less than 0.1 mg of chromium and selenium.
Amplification Figure 1.27 Possible
How can elements that are present in such minuscule Concentrations of an Essential Element in the Diet amounts have such large effects on an organism’s health? Our knowledge of the pathways by which each of the known trace elements affects health is far from complete, but certain general features are clear. The trace elements participate in an amplification mechanism; that is, they are essential components of larger biological molecules that are capable of interacting with or regulating the levels of relatively The deficient, optimum, and large amounts of other molecules. For example, vitamin toxic concentrations are different for different elements. B12 contains a single atom of cobalt, which is essential for its biological function. If the molecule whose level is controlled by the trace element can regulate the level of another molecule, and more and more molecules, then the potential exists for extreme amplification of small variations in the level of the trace element. One goal of modern chemical research is to elucidate in detail the roles of the essential elements. In subsequent chapters, we will introduce some results of this research to demonstrate the biological importance of many of the elements and their compounds.
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Summary About 19 of the approximately 115 known elements are essential for humans. An essential element is one whose absence results in abnormal biological function or development that is prevented by dietary supplementation with that element. Living organisms contain relatively large amounts of oxygen, carbon, hydrogen, nitrogen, and sulfur (these five elements are known as the bulk elements), along with sodium, magnesium, potassium, calcium, chlorine, and phosphorus (these six elements are known as macrominerals). The other essential elements are the trace elements, which are present in very small quantities. Dietary intakes of elements range from deficient to optimum to toxic with increasing quantities; the optimum levels differ greatly for the essential elements.
KEY TAKEAWAY • The absence of some elements can result in abnormal biological function or development.
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1.9 Essential Skills 1 TOPICS • • • •
Measurement Scientific Notation Significant Figures Accuracy and Precision
This section describes some of the fundamental mathematical skills you will need to complete the questions and problems in this text. For some of you, this discussion will serve as a review, whereas others may be encountering at least some of the ideas and techniques for the first time. We will introduce other mathematical skills in subsequent Essential Skills sections as the need arises. Be sure you are familiar with the topics discussed here before you start the Chapter 1 "Introduction to Chemistry" problems.
Measurement Instruments of Measurement
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Graduated glassware is used to deliver variable volumes of liquid.
Volumetric glassware is used to deliver (pipette) or contain (volumetric flask) a single volume accurately when filled to the calibration mark.
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A balance is used to measure mass.
A variety of instruments are available for making direct measurements of the macroscopic properties of a chemical substance. For example, we usually measure the volume62 of a liquid sample with pipettes, burets, graduated cylinders, and volumetric flasks, whereas we usually measure the mass of a solid or liquid substance with a balance. Measurements on an atomic or molecular scale, in contrast, require specialized instrumentation, such as the mass spectrometer described in Section 1.6 "Isotopes and Atomic Masses".
SI Units
62. The amount of space occupied by a sample of matter. 63. A system of units based on metric units that requires measurements to be expressed in decimal form. There are seven base units in the SI system.
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All reported measurements must include an appropriate unit of measurement because to say that a substance has “a mass of 10,” for example, does not tell whether the mass was measured in grams, pounds, tons, or some other unit. To establish worldwide standards for the consistent measurement of important physical and chemical properties, an international body called the General Conference on Weights and Measures devised the Système internationale d’unités (or SI)63. The International System of Units is based on metric units and requires that measurements be expressed in decimal form. Table 1.7 "SI Base Units" lists the seven base units of the SI system; all other SI units of measurement are derived from them.
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By attaching prefixes to the base unit, the magnitude of the unit is indicated; each prefix indicates that the base unit is multiplied by a specified power of 10. The prefixes, their symbols, and their numerical significance are given in Table 1.8 "Prefixes Used with SI Units". To study chemistry, you need to know the information presented in Table 1.7 "SI Base Units" and Table 1.8 "Prefixes Used with SI Units". Table 1.7 SI Base Units Base Quantity
Unit Name Abbreviation
mass
kilogram
kg
length
meter
m
time
second
s
temperature
kelvin
K
electric current
ampere
A
amount of substance mole
mol
luminous intensity
cd
candela
Table 1.8 Prefixes Used with SI Units Prefix Symbol
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Value
Power of 10
Meaning
tera
T
1,000,000,000,000
1012
trillion
giga
G
1,000,000,000
109
billion
mega
M
1,000,000
106
million
kilo
k
1000
103
thousand
hecto
h
100
102
hundred
deca
da
10
101
ten
—
—
1
100
one
deci
d
0.1
10−1
tenth
centi
c
0.01
10−2
hundredth
milli
m
0.001
10−3
thousandth
micro
μ
0.000001
10−6
millionth
nano
n
0.000000001
10−9
billionth
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Prefix Symbol
Value
Power of 10
Meaning
pico
p
0.000000000001
10−12
trillionth
femto
f
0.000000000000001
10−15
quadrillionth
Units of Mass, Volume, and Length The units of measurement you will encounter most frequently in chemistry are those for mass, volume, and length. The basic SI unit for mass is the kilogram (kg), but in the laboratory, mass is usually expressed in either grams (g) or milligrams (mg): 1000 g = 1 kg, 1000 mg = 1 g, and 1,000,000 mg = 1 kg. Units for volume are derived from the cube of the SI unit for length, which is the meter (m). Thus the basic SI unit for volume is cubic meters (length × width × height = m3). In chemistry, however, volumes are usually reported in cubic centimeters (cm 3) and cubic decimeters (dm3) or milliliters (mL) and liters (L), although the liter is not an SI unit of measurement. The relationships between these units are as follows: 1 L = 1000 mL = 1 dm3 1 mL = 1 cm3 1000 cm3 = 1 L
Scientific Notation Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator requires a display with at least 24 decimal places. A system called scientific notation64 avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form N × 10n
64. A system that expresses numbers in the form N × 10n, where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10) and n is an integer that can be either positive or negative (100 = 1). The purpose of scientific notation is to simplify the manipulation of numbers with large or small magnitudes.
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where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (100 = 1). The number 10 is called the base because it is this number that is raised to the power n. Although a base number may have values other than 10, the base number in scientific notation is always 10. A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:
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• If the decimal point is moved to the left n places, n is positive. • If the decimal point is moved to the right n places, n is negative. Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Skill Builder ES1.
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SKILL BUILDER ES1 Convert each number to scientific notation. a. b. c. d. e. f. g. h.
637.8 0.0479 7.86 12,378 0.00032 61.06700 2002.080 0.01020
Solution
a. To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left:
63 7 .8 ↖↖
Because the decimal point was moved two places to the left, n = 2. In scientific notation, 637.8 = 6.378 × 102. b. To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right:
0.0 4 79 ↗↗
Because the decimal point was moved two places to the right, n = −2. In scientific notation, 0.0479 = 4.79 × 10−2. c. 7.86 × 100: this is usually expressed simply as 7.86. (Recall that 10 0 = 1.) d. 1.2378 × 104; because the decimal point was moved four places to the left, n = 4. e. 3.2 × 10−4; because the decimal point was moved four places to the right, n = −4. f. 6.106700 × 101: this is usually expressed as 6.1067 × 10. g. 2.002080 × 103 h. 1.020 × 10−2
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Addition and Subtraction Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Skill Builder ES2 illustrates how to do this.
SKILL BUILDER ES2 Carry out the appropriate operation on each number and then express the answer in scientific notation. a. (1.36 × 102) + (4.73 × 103) b. (6.923 × 10−3) − (8.756 × 10−4) Solution
a. Both exponents must have the same value, so these numbers are converted to either (1.36 × 102) + (47.3 × 102) or (0.136 × 103) + (4.73 × 103). Choosing either alternative gives the same answer, reported to two decimal places: (1.36 × 102) + (47.3 × 102) = (1.36 + 47.3) × 102 = 48.66 × 102 = 4.87 × 103 (0.136 × 103) + (4.73 × 103) = (0.136 + 4.73) × 103 = 4.87 × 103 In converting 48.66 × 102 to scientific notation, n has become more positive by 1 because the value of N has decreased. b. Converting the exponents to the same value gives either (6.923 × 10−3) − (0.8756 × 10−3) or (69.23 × 10−4) − (8.756 × 10−4). Completing the calculations gives the same answer, expressed to three decimal places: (6.923 × 10−3) − (0.8756 × 10−3) = (6.923 − 0.8756) × 10−3 = 6.047 × 10−3 (69.23 × 10−4) − (8.756 × 10−4) = (69.23 − 8.756) × 10−4 = 60.474 × 10−4 = 6.047 × 10−3
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Multiplication and Division When multiplying numbers expressed in scientific notation, we multiply the values of N and add together the values of n. Conversely, when dividing, we divide N in the dividend (the number being divided) by N in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Skill Builder ES3.
SKILL BUILDER ES3 Perform the appropriate operation on each expression and express your answer in scientific notation. a. (6.022 × 1023)(6.42 × 10−2) b. c.
1.67×10 −24 9.12×10 −28 (6.63×10 −34 )(6.0×10) 8.52×10 −2
Solution
a. In multiplication, we add the exponents:
(6.022 × 10 23 )(6. 42 × 10 −2 ) = (6.022)(6.42) × 10 [23 + (−2)] = 3 b. In division, we subtract the exponents:
1.67 × 10 −24 9.12 × 10 −28
=
1.67 × 10 [−24 − (−28)] = 0.183 × 10 4 = 1.83 × 10 3 9.12
c. This problem has both multiplication and division:
(6.63 × 10 −34 )(6.0 × 10) (8.52 × 10 −2 )
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=
39.78 × 10 [−34 + 1 − (−2)] = 4.7 × 10 −3 8.52
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Significant Figures No measurement is free from error. Error is introduced by (1) the limitations of instruments and measuring devices (such as the size of the divisions on a graduated cylinder) and (2) the imperfection of human senses. Although errors in calculations can be enormous, they do not contribute to uncertainty in measurements. Chemists describe the estimated degree of error in a measurement as the uncertainty65 of the measurement, and they are careful to report all measured values using only significant figures66, numbers that describe the value without exaggerating the degree to which it is known to be accurate. Chemists report as significant all numbers known with absolute certainty, plus one more digit that is understood to contain some uncertainty. The uncertainty in the final digit is usually assumed to be ±1, unless otherwise stated. The following rules have been developed for counting the number of significant figures in a measurement or calculation:
65. The estimated degree of error in a measurement. The degree of uncertainty in a measurement can be indicated by reporting all significant figures plus one. 66. Numbers that describe the value without exaggerating the degree to which it is known to be accurate. 67. An integer obtained either by counting objects or from definitions (e.g., 1 in. = 2.54 cm). Exact numbers have infinitely many significant figures.
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1. Any nonzero digit is significant. 2. Any zeros between nonzero digits are significant. The number 2005, for example, has four significant figures. 3. Any zeros used as a placeholder preceding the first nonzero digit are not significant. So 0.05 has one significant figure because the zeros are used to indicate the placement of the digit 5. In contrast, 0.050 has two significant figures because the last two digits correspond to the number 50; the last zero is not a placeholder. As an additional example, 5.0 has two significant figures because the zero is used not to place the 5 but to indicate 5.0. 4. When a number does not contain a decimal point, zeros added after a nonzero number may or may not be significant. An example is the number 100, which may be interpreted as having one, two, or three significant figures. (Note: treat all trailing zeros in exercises and problems in this text as significant unless you are specifically told otherwise.) 5. Integers obtained either by counting objects or from definitions are exact numbers67, which are considered to have infinitely many significant figures. If we have counted four objects, for example, then the number 4 has an infinite number of significant figures (i.e., it represents 4.000…). Similarly, 1 foot (ft) is defined to contain 12 inches (in), so the number 12 in the following equation has infinitely many significant figures: 1 ft = 12 in
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An effective method for determining the number of significant figures is to convert the measured or calculated value to scientific notation because any zero used as a placeholder is eliminated in the conversion. When 0.0800 is expressed in scientific notation as 8.00 × 10−2, it is more readily apparent that the number has three significant figures rather than five; in scientific notation, the number preceding the exponential (i.e., N) determines the number of significant figures. Skill Builder ES4 provides practice with these rules.
SKILL BUILDER ES4 Give the number of significant figures in each. Identify the rule for each. a. b. c. d. e. f.
5.87 0.031 52.90 00.2001 500 6 atoms
Solution a. three (rule 1) b. two (rule 3); in scientific notation, this number is represented as 3.1 × 10−2, showing that it has two significant figures. c. four (rule 3) d. four (rule 2); this number is 2.001 × 10−1 in scientific notation, showing that it has four significant figures. e. one, two, or three (rule 4) f. infinite (rule 5)
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SKILL BUILDER ES5 Which measuring apparatus would you use to deliver 9.7 mL of water as accurately as possible? To how many significant figures can you measure that volume of water with the apparatus you selected?
Solution Use the 10 mL graduated cylinder, which will be accurate to two significant figures.
Mathematical operations are carried out using all the digits given and then rounding the final result to the correct number of significant figures to obtain a reasonable answer. This method avoids compounding inaccuracies by successively rounding intermediate calculations. After you complete a calculation, you may have to round the last significant figure up or down depending on the value of the digit that follows it. If the digit is 5 or greater, then the number is rounded up. For example, when rounded to three significant figures, 5.215 is 5.22, whereas 5.213 is 5.21. Similarly, to three significant figures, 5.005 kg becomes 5.01 kg, whereas 5.004 kg becomes 5.00 kg. The procedures for dealing with significant figures are different for addition and subtraction versus multiplication and division.
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When we add or subtract measured values, the value with the fewest significant figures to the right of the decimal point determines the number of significant figures to the right of the decimal point in the answer. Drawing a vertical line to the right of the column corresponding to the smallest number of significant figures is a simple method of determining the proper number of significant figures for the answer:
+
3240.7 21.2 36 3261.9 36
The line indicates that the digits 3 and 6 are not significant in the answer. These digits are not significant because the values for the corresponding places in the other measurement are unknown (3240.7??). Consequently, the answer is expressed as 3261.9, with five significant figures. Again, numbers greater than or equal to 5 are rounded up. If our second number in the calculation had been 21.256, then we would have rounded 3261.956 to 3262.0 to complete our calculation. When we multiply or divide measured values, the answer is limited to the smallest number of significant figures in the calculation; thus, 42.9 × 8.323 = 357.057 = 357. Although the second number in the calculation has four significant figures, we are justified in reporting the answer to only three significant figures because the first number in the calculation has only three significant figures. An exception to this rule occurs when multiplying a number by an integer, as in 12.793 × 12. In this case, the number of significant figures in the answer is determined by the number 12.973, because we are in essence adding 12.973 to itself 12 times. The correct answer is therefore 155.516, an increase of one significant figure, not 155.52. When you use a calculator, it is important to remember that the number shown in the calculator display often shows more digits than can be reported as significant in your answer. When a measurement reported as 5.0 kg is divided by 3.0 L, for example, the display may show 1.666666667 as the answer. We are justified in reporting the answer to only two significant figures, giving 1.7 kg/L as the answer, with the last digit understood to have some uncertainty. In calculations involving several steps, slightly different answers can be obtained depending on how rounding is handled, specifically whether rounding is performed on intermediate results or postponed until the last step. Rounding to the correct number of significant figures should always be performed at the end of a series of calculations because rounding of intermediate results can sometimes cause the final answer to be significantly in error.
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In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate roundings need to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end. In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step. Skill Builder ES6 provides practice with calculations using significant figures.
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SKILL BUILDER ES6 Complete the calculations and report your answers using the correct number of significant figures. a. b. c. d. e. f. g. h. i.
87.25 mL + 3.0201 mL 26.843 g + 12.23 g 6 × 12.011 2(1.008) g + 15.99 g 137.3 + 2(35.45) 118.7 2
g − 35. 5 g 47. 23 g − 207.2 g 5.92 77.604 − 4. 8 6.467 24.86 − 3. 26(0.98) 2.0
j. (15.9994 × 9) + 2.0158 Solution a. b. c. d. e. f. g. h. i. j.
90.27 mL 39.07 g 72.066 (See rule 5 under “Significant Figures.”) 2(1.008) g + 15.99 g = 2.016 g + 15.99 g = 18.01 g 137.3 + 2(35.45) = 137.3 + 70.90 = 208.2 59.35 g − 35.5 g = 23.9 g 47.23 g − 35.0 g = 12.2 g 12.00 − 4.8 = 7.2 12 − 3.2 = 9 143.9946 + 2.0158 = 146.0104
Accuracy and Precision Measurements may be accurate68, meaning that the measured value is the same as the true value; they may be precise69, meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise. 68. The measured value is the same as the true value. 69. Multiple measurements give nearly identical values.
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Suppose, for example, that the mass of a sample of gold was measured on one balance and found to be 1.896 g. On a different balance, the same sample was found to have a mass of 1.125 g. Which was correct? Careful and repeated measurements,
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including measurements on a calibrated third balance, showed the sample to have a mass of 1.895 g. The masses obtained from the three balances are in the following table: Balance 1 Balance 2 Balance 3 1.896 g
1.125 g
1.893 g
1.895 g
1.158 g
1.895 g
1.894 g
1.067 g
1.895 g
Whereas the measurements obtained from balances 1 and 3 are reproducible (precise) and are close to the accepted value (accurate), those obtained from balance 2 are neither. Even if the measurements obtained from balance 2 had been precise (if, for example, they had been 1.125, 1.124, and 1.125), they still would not have been accurate. We can assess the precision of a set of measurements by calculating the average deviation of the measurements as follows: 1. Calculate the average value of all the measurements:
average =
sum of measurements number of measurements
2. Calculate the deviation of each measurement, which is the absolute value of the difference between each measurement and the average value: deviation = |measurement − average| where | | means absolute value (i.e., convert any negative number to a positive number). 3. Add all the deviations and divide by the number of measurements to obtain the average deviation:
average =
sum of deviations number of measurements
Then we can express the precision as a percentage by dividing the average deviation by the average value of the measurements and multiplying the result by 100. In the case of balance 2, the average value is
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1.125 g + 1.158 + 1.067 g = 1.117 g 3 The deviations are 1.125 g − 1.117 g = 0.008 g, 1.158 g − 1.117 g = 0.041 g, and |1.067 g − 1.117 g| = 0.050 g. So the average deviation is
0.008 g + 0.041 g + 0.050 g = 0.033 g 3 The precision of this set of measurements is therefore
0.033 g 1.117 g
× 100 = 3.0%
When a series of measurements is precise but not accurate, the error is usually systematic. Systematic errors can be caused by faulty instrumentation or faulty technique. The difference between accuracy and precision is demonstrated in Skill Builder ES7.
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SKILL BUILDER ES7 The following archery targets show marks that represent the results of four sets of measurements. Which target shows a. b. c. d.
a precise but inaccurate set of measurements? an accurate but imprecise set of measurements? a set of measurements that is both precise and accurate? a set of measurements that is neither precise nor accurate?
Solution a. b. c. d.
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(c) (a) (b) (d)
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SKILL BUILDER ES8 a. A 1-carat diamond has a mass of 200.0 mg. When a jeweler repeatedly weighed a 2-carat diamond, he obtained measurements of 450.0 mg, 459.0 mg, and 463.0 mg. Were the jeweler’s measurements accurate? Were they precise? b. A single copper penny was tested three times to determine its composition. The first analysis gave a composition of 93.2% zinc and 2.8% copper, the second gave 92.9% zinc and 3.1% copper, and the third gave 93.5% zinc and 2.5% copper. The actual composition of the penny was 97.6% zinc and 2.4% copper. Were the results accurate? Were they precise? Solution
a. The expected mass of a 2-carat diamond is 2 × 200.0 mg = 400.0 mg. The average of the three measurements is 457.3 mg, about 13% greater than the true mass. These measurements are not particularly accurate. The deviations of the measurements are 7.3 mg, 1.7 mg, and 5.7 mg, respectively, which give an average deviation of 4.9 mg and a precision of
4.9 mg 457.3 mg
× 100 = 1.1%
These measurements are rather precise. b. The average values of the measurements are 93.2% zinc and 2.8% copper versus the true values of 97.6% zinc and 2.4% copper. Thus these measurements are not very accurate, with errors of −4.5% and + 17% for zinc and copper, respectively. (The sum of the measured zinc and copper contents is only 96.0% rather than 100%, which tells us that either there is a significant error in one or both measurements or some other element is present.) The deviations of the measurements are 0.0%, 0.3%, and 0.3% for both zinc and copper, which give an average deviation of 0.2% for
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both metals. We might therefore conclude that the measurements are equally precise, but that is not the case. Recall that precision is the average deviation divided by the average value times 100. Because the average value of the zinc measurements is much greater than the average value of the copper measurements (93.2% versus 2.8%), the copper measurements are much less precise.
0.2% × 100 = 0.2% 93.2% 0.2% precision (Cu) = × 100 = 7% 2.8% precision (Zn) =
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APPLICATION PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts. 1. In 1953, James Watson and Francis Crick spent three days analyzing data to develop a model that was consistent with the known facts about the structure of DNA, the chemical substance that is the basis for life. They were awarded the Nobel Prize in Physiology or Medicine for their work. Based on this information, would you classify their proposed model for the structure of DNA as an experiment, a law, a hypothesis, or a theory? Explain your reasoning. 2. In each scenario, state the observation and the hypothesis. a. A recently discovered Neanderthal throat bone has been found to be similar in dimensions and appearance to that of modern humans; therefore, some scientists believe that Neanderthals could talk. b. Because DNA profiles from samples of human tissue are widely used in criminal trials, DNA sequences from plant residue on clothing can be used to place a person at the scene of a crime. 3. Small quantities of gold from far underground are carried to the surface by groundwater, where the gold can be taken up by certain plants and stored in their leaves. By identifying the kinds of plants that grow around existing gold deposits, one should be able to use this information to discover potential new gold deposits. a. State the observation. b. State the hypothesis. c. Devise an experiment to test the hypothesis. 4. Large amounts of nitrogen are used by the electronics industry to provide a gas blanket over a component during production. This ensures that undesired reactions with oxygen will not occur. Classify each statement as an extensive property or an intensive property of nitrogen. a. b. c. d.
Nitrogen is a colorless gas. A volume of 22.4 L of nitrogen gas weighs 28 g at 0°C. Liquid nitrogen boils at 77.4 K. Nitrogen gas has a density of 1.25 g/L at 0°C.
5. Oxygen is the third most abundant element in the universe and makes up about two-thirds of the human body. Classify each statement as an extensive property or an intensive property of oxygen.
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a. b. c. d.
Liquid oxygen boils at 90.2 K. Liquid oxygen is pale blue. A volume of 22.4 L of oxygen gas weighs 32 g at 0°C. Oxygen has a density of 1.43 g/L at 0°C.
6. One of the first high-temperature superconductors was found to contain elements in the ratio 1Y:2Ba:3Cu:6.8O. A material that contains elements in the ratio 1Y:2Ba:3Cu:6O, however, was not a high-temperature superconductor. Do these materials obey the law of multiple proportions? Is the ratio of elements in each compound consistent with Dalton’s law of indivisible atoms? 7. ♦ There has been increased evidence that human activities are causing changes in Earth’s atmospheric chemistry. Recent research efforts have focused on atmospheric ozone (O3) concentrations. The amount of ozone in the atmosphere is influenced by concentrations of gases that contain only nitrogen and oxygen, among others. The following table gives the masses of nitrogen that combine with 1.00 g of oxygen to form three of these compounds. Compound Mass of Nitrogen (g) A
0.875
B
0.438
C
0.350
a. Determine the ratios of the masses of nitrogen that combine with 1.00 g of oxygen in these compounds. Are these data consistent with the law of multiple proportions? b. Predict the mass of nitrogen that would combine with 1.00 g of oxygen to form another possible compound in the series. 8. Indium has an average atomic mass of 114.818 amu. One of its two isotopes has an atomic mass of 114.903 amu with a percent abundance of 95.70. What is the mass of the other isotope? 9. Earth’s core is largely composed of iron, an element that is also a major component of black sands on beaches. Iron has four stable isotopes. Use the data to calculate the average atomic mass of iron. Isotope Percent Abundance (%) Atomic Mass (amu)
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54Fe
5.82
53.9396
56Fe
91.66
55.9349
57Fe
2.19
56.9354
58Fe
0.33
57.9333
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10. ♦ Because ores are deposited during different geologic periods, lead ores from different mining regions of the world can contain different ratios of isotopes. Archaeologists use these differences to determine the origin of lead artifacts. For example, the following table lists the percent abundances of three lead isotopes from one artifact recovered from Rio Tinto in Spain. Isotope Percent Abundance (%) Atomic mass (amu) 204Pb
203.973028
206Pb
24.41
205.974449
207Pb
20.32
206.97580
208Pb
50.28
207.976636
a. If the only other lead isotope in the artifact is 204Pb, what is its percent abundance? b. What is the average atomic mass of lead if the only other isotope in the artifact is 204Pb? c. An artifact from Laurion, Greece, was found to have a 207Pb:206Pb ratio of 0.8307. From the data given, can you determine whether the lead in the artifact from Rio Tinto came from the same source as the lead in the artifact from Laurion, Greece? 11. The macrominerals sodium, magnesium, potassium, calcium, chlorine, and phosphorus are widely distributed in biological substances, although their distributions are far from uniform. Classify these elements by both their periods and their groups and then state whether each is a metal, a nonmetal, or a semimetal. If a metal, is the element a transition metal? 12. The composition of fingernails is sensitive to exposure to certain elements, including sodium, magnesium, aluminum, chlorine, potassium, calcium, selenium, vanadium, chromium, manganese, iron, cobalt, copper, zinc, scandium, arsenic, and antimony. Classify these elements by both their periods and their groups and then determine whether each is a metal, a nonmetal, or a semimetal. Of the metals, which are transition metals? Based on your classifications, predict other elements that could prove to be detectable in fingernails. 13. Mercury levels in hair have been used to identify individuals who have been exposed to toxic levels of mercury. Is mercury an essential element? a trace element? 14. Trace elements are usually present at levels of less than 50 mg/kg of body weight. Classify the essential trace elements by their groups and periods in the
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periodic table. Based on your classifications, would you predict that arsenic, cadmium, and lead are potential essential trace elements?
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Chapter 2 Molecules, Ions, and Chemical Formulas Chapter 1 "Introduction to Chemistry" introduced some of the fundamental concepts of chemistry, with particular attention to the basic properties of atoms and elements. These entities are the building blocks of all substances we encounter, yet most common substances do not consist of only pure elements or individual atoms. Instead, nearly all substances are chemical compounds or mixtures of chemical compounds. Although there are only about 115 elements (of which about 86 occur naturally), millions of chemical compounds are known, with a tremendous range of physical and chemical properties. Consequently, the emphasis of modern chemistry (and this text) is on understanding the relationship between the structures and properties of chemical compounds.
Petroleum refining. Using chemicals, catalysts, heat, and pressure, a petroleum refinery will separate, combine, and rearrange the structure and bonding patterns of the basic carbon-hydrogen molecules found in crude oil. The final products include gasoline, paraffin, diesel fuel, lubricants, and bitumen.
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In this chapter, you will learn how to describe the composition of chemical compounds. We introduce you to chemical nomenclature—the language of chemistry—that will enable you to recognize and name the most common kinds of compounds. An understanding of chemical nomenclature not only is essential for your study of chemistry but also has other benefits—for example, it helps you understand the labels on products found in the supermarket and the pharmacy. You will also be better equipped to understand many of the important environmental and medical issues that face society. By the end of this chapter, you will be able to describe what happens chemically when a doctor prepares a cast to stabilize a broken bone, and you will know the composition of common substances such as laundry bleach, the active ingredient in baking powder, and the foulsmelling compound responsible for the odor of spoiled fish. Finally, you will be able to explain the chemical differences among different grades of gasoline.
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2.1 Chemical Compounds LEARNING OBJECTIVE 1. To understand the differences between covalent and ionic bonding.
1. An interaction between electrically charged particles such as protons and electrons. 2. An electrostatic interaction between oppositely charged species (positive and negative) that results in a force that causes them to move toward each other. 3. An electrostatic interaction between two species that have the same charge (both positive or both negative) that results in a force that causes them to repel each other. 4. An attractive interaction between atoms that holds them together in compounds. 5. A compound consisting of positively charged ions (cations) and negatively charged ions (anions) held together by strong electrostatic forces. 6. A compound that consists of discrete molecules. 7. A group of atoms in which one or more pairs of electrons are shared between bonded atoms.
The atoms in all substances that contain more than one atom are held together by electrostatic interactions1—interactions between electrically charged particles such as protons and electrons. Electrostatic attraction2 between oppositely charged species (positive and negative) results in a force that causes them to move toward each other, like the attraction between opposite poles of two magnets. In contrast, electrostatic repulsion3 between two species with the same charge (either both positive or both negative) results in a force that causes them to repel each other, as do the same poles of two magnets. Atoms form chemical compounds when the attractive electrostatic interactions between them are stronger than the repulsive interactions. Collectively, we refer to the attractive interactions between atoms as chemical bonds4. Chemical bonds are generally divided into two fundamentally different kinds: ionic and covalent. In reality, however, the bonds in most substances are neither purely ionic nor purely covalent, but they are closer to one of these extremes. Although purely ionic and purely covalent bonds represent extreme cases that are seldom encountered in anything but very simple substances, a brief discussion of these two extremes helps us understand why substances that have different kinds of chemical bonds have very different properties. Ionic compounds5 consist of positively and negatively charged ions held together by strong electrostatic forces, whereas covalent compounds6 generally consist of molecules7, which are groups of atoms in which one or more pairs of electrons are shared between bonded atoms. In a covalent bond8, the atoms are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share. We begin our discussion of structures and formulas by describing covalent compounds. The energetic factors involved in bond formation are described in more quantitative detail in Chapter 8 "Ionic versus Covalent Bonding".
8. The electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons they share.
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Note the Pattern Ionic compounds consist of ions of opposite charges held together by strong electrostatic forces, whereas pairs of electrons are shared between bonded atoms in covalent compounds.
Covalent Molecules and Compounds Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (“two atoms”) molecules H2, N2, O2, F2, Cl2, Br2, and I2 (part (a) in Figure 2.1 "Elements That Exist as Covalent Molecules"). Similarly, a few pure elements are polyatomic9 (“many atoms”) molecules, such as elemental phosphorus and sulfur, which occur as P4 and S8 (part (b) in Figure 2.1 "Elements That Exist as Covalent Molecules"). Each covalent compound is represented by a molecular formula10, which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as H2O. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as CO2. Figure 2.1 Elements That Exist as Covalent Molecules
9. Molecules that contain more than two atoms. 10. A representation of a covalent compound that consists of the atomic symbol for each component element (in a prescribed order) accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number is greater than 1.
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(a) Several elements naturally exist as diatomic molecules, in which two atoms (E) are joined by one or more covalent bonds to form a molecule with the general formula E2. (b) A few elements naturally exist as polyatomic molecules, which contain more than two atoms. For example, phosphorus exists as P 4 tetrahedra—regular polyhedra with four triangular sides—with a phosphorus atom at each vertex. Elemental sulfur consists of a puckered ring of eight sulfur atoms connected by single bonds. Selenium is not shown due to the complexity of its structure.
Covalent compounds that contain predominantly carbon and hydrogen are called organic compounds11. The convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH4O is methyl alcohol, a fuel). Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compounds12; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table (see Chapter 32 "Appendix H: Periodic Table of Elements"), such as we see in CO2 or SF6. Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13–15, the hydrogen is usually listed last in the formula. Examples are ammonia (NH3) and silane (SiH4). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H2O, not OH2. The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H 2SO4), are described in Section 2.5 "Acids and Bases".
Note the Pattern For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up.
11. A covalent compound that contains predominantly carbon and hydrogen. 12. An ionic or covalent compound that consists primarily of elements other than carbon and hydrogen.
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EXAMPLE 1 Write the molecular formula of each compound. a. The phosphorus-sulfur compound that is responsible for the ignition of so-called strike anywhere matches has 4 phosphorus atoms and 3 sulfur atoms per molecule. b. Ethyl alcohol, the alcohol of alcoholic beverages, has 1 oxygen atom, 2 carbon atoms, and 6 hydrogen atoms per molecule. c. Freon-11, once widely used in automobile air conditioners and implicated in damage to the ozone layer, has 1 carbon atom, 3 chlorine atoms, and 1 fluorine atom per molecule. Given: identity of elements present and number of atoms of each Asked for: molecular formula Strategy: A Identify the symbol for each element in the molecule. Then identify the substance as either an organic compound or an inorganic compound. B If the substance is an organic compound, arrange the elements in order beginning with carbon and hydrogen and then list the other elements alphabetically. If it is an inorganic compound, list the elements beginning with the one farthest left in the periodic table. List elements in the same group starting with the lower element and working up. C From the information given, add a subscript for each kind of atom to write the molecular formula. Solution: a. A The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic. B Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first. C Writing the number of each kind of atom as a right-hand subscript gives P 4S3 as the molecular formula. b. A Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound. B The formula for an organic compound is written
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with the number of carbon atoms first, the number of hydrogen atoms next, and the other atoms in alphabetical order: CHO. C Adding subscripts gives the molecular formula C2H6O. c. A Freon-11 contains carbon, chlorine, and fluorine. It can be viewed as either an inorganic compound or an organic compound (in which fluorine has replaced hydrogen). The formula for Freon-11 can therefore be written using either of the two conventions. B According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl3F. C We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl 3F. Exercise Write the molecular formula for each compound. a. Nitrous oxide, also called “laughing gas,” has 2 nitrogen atoms and 1 oxygen atom per molecule. Nitrous oxide is used as a mild anesthetic for minor surgery and as the propellant in cans of whipped cream. b. Sucrose, also known as cane sugar, has 12 carbon atoms, 11 oxygen atoms, and 22 hydrogen atoms. c. Sulfur hexafluoride, a gas used to pressurize “unpressurized” tennis balls and as a coolant in nuclear reactors, has 6 fluorine atoms and 1 sulfur atom per molecule. Answer: a. N2O b. C12H22O11 c. SF6
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Representations of Molecular Structures Molecular formulas give only the elemental composition of molecules. In contrast, structural formulas13 show which atoms are bonded to one another and, in some cases, the approximate arrangement of the atoms in space. Knowing the structural formula of a compound enables chemists to create a three-dimensional model, which provides information about how that compound will behave physically and chemically. The structural formula for H2 can be drawn as H–H and that for I2 as I–I, where the line indicates a single pair of shared electrons, a single bond14. Two pairs of electrons are shared in a double bond15, which is indicated by two lines— for example, O2 is O=O. Three electron pairs are shared in a triple bond16, which is indicated by three lines—for example, N2 is N≡N (see Figure 2.2 "Molecules That Contain Single, Double, and Triple Bonds"). Carbon is unique in the extent to which it forms single, double, and triple bonds to itself and other elements. The number of bonds formed by an atom in its covalent compounds is not arbitrary. As you will learn in Chapter 8 "Ionic versus Covalent Bonding", hydrogen, oxygen, nitrogen, and carbon have a very strong tendency to form substances in which they have one, two, three, and four bonds to other atoms, respectively (Table 2.1 "The Number of Bonds That Selected Atoms Commonly Form to Other Atoms"). Figure 2.2 Molecules That Contain Single, Double, and Triple Bonds
13. A representation of a molecule that shows which atoms are bonded to one another and, in some cases, the approximate arrangement of atoms in space. 14. A chemical bond formed when two atoms share a single pair of electrons. 15. A chemical bond formed when two atoms share two pairs of electrons. 16. A chemical bond formed when two atoms share three pairs of electrons.
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Hydrogen (H2) has a single bond between atoms. Oxygen (O2) has a double bond between atoms, indicated by two lines (=). Nitrogen (N2) has a triple bond between atoms, indicated by three lines (≡). Each bond represents an electron pair.
Table 2.1 The Number of Bonds That Selected Atoms Commonly Form to Other Atoms Atom
Number of Bonds
H (group 1)
1
O (group 16)
2
N (group 15)
3
C (group 14)
4
The structural formula for water can be drawn as follows:
Because the latter approximates the experimentally determined shape of the water molecule, it is more informative. Similarly, ammonia (NH 3) and methane (CH4) are often written as planar molecules:
As shown in Figure 2.3 "The Three-Dimensional Structures of Water, Ammonia, and Methane", however, the actual three-dimensional structure of NH3 looks like a pyramid with a triangular base of three hydrogen atoms. The structure of CH 4, with four hydrogen atoms arranged around a central carbon atom as shown in Figure 2.3 "The Three-Dimensional Structures of Water, Ammonia, and Methane", is tetrahedral. That is, the hydrogen atoms are positioned at every other vertex of a cube. Many compounds—carbon compounds, in particular—have four bonded atoms arranged around a central atom to form a tetrahedron.
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Figure 2.3 The Three-Dimensional Structures of Water, Ammonia, and Methane
(a) Water is a V-shaped molecule, in which all three atoms lie in a plane. (b) In contrast, ammonia has a pyramidal structure, in which the three hydrogen atoms form the base of the pyramid and the nitrogen atom is at the vertex. (c) The four hydrogen atoms of methane form a tetrahedron; the carbon atom lies in the center.
Figure 2.1 "Elements That Exist as Covalent Molecules", Figure 2.2 "Molecules That Contain Single, Double, and Triple Bonds", and Figure 2.3 "The Three-Dimensional Structures of Water, Ammonia, and Methane" illustrate different ways to represent the structures of molecules. It should be clear that there is no single “best” way to draw the structure of a molecule; the method you use depends on which aspect of the structure you want to emphasize and how much time and effort you want to spend. Figure 2.4 "Different Ways of Representing the Structure of a Molecule" shows some of the different ways to portray the structure of a slightly more complex CH4. Methane has a threemolecule: methanol. These representations differ dimensional, tetrahedral greatly in their information content. For example, the structure. molecular formula for methanol (part (a) in Figure 2.4 "Different Ways of Representing the Structure of a Molecule") gives only the number of each kind of atom; writing methanol as CH4O tells nothing about its structure. In contrast, the structural formula (part (b) in Figure 2.4 "Different Ways of Representing the Structure of a Molecule") indicates how the atoms are connected, but it makes methanol look as if it is planar (which it is not). Both the ball-and-stick model (part (c) in Figure 2.4 "Different Ways of Representing the Structure of a Molecule") and the perspective drawing (part (d) in Figure 2.4 "Different Ways of Representing the Structure of a Molecule") show the three-
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dimensional structure of the molecule. The latter (also called a wedge-and-dash representation) is the easiest way to sketch the structure of a molecule in three dimensions. It shows which atoms are above and below the plane of the paper by using wedges and dashes, respectively; the central atom is always assumed to be in the plane of the paper. The space-filling model (part (e) in Figure 2.4 "Different Ways of Representing the Structure of a Molecule") illustrates the approximate relative sizes of the atoms in the molecule, but it does not show the bonds between the atoms. Also, in a space-filling model, atoms at the “front” of the molecule may obscure atoms at the “back.” Figure 2.4 Different Ways of Representing the Structure of a Molecule
(a) The molecular formula for methanol gives only the number of each kind of atom present. (b) The structural formula shows which atoms are connected. (c) The ball-and-stick model shows the atoms as spheres and the bonds as sticks. (d) A perspective drawing (also called a wedge-and-dash representation) attempts to show the threedimensional structure of the molecule. (e) The space-filling model shows the atoms in the molecule but not the bonds. (f) The condensed structural formula is by far the easiest and most common way to represent a molecule.
Although a structural formula, a ball-and-stick model, a perspective drawing, and a space-filling model provide a significant amount of information about the structure of a molecule, each requires time and effort. Consequently, chemists often use a condensed structural formula (part (f) in Figure 2.4 "Different Ways of Representing the Structure of a Molecule"), which omits the lines representing bonds between atoms and simply lists the atoms bonded to a given atom next to it. Multiple groups attached to the same atom are shown in parentheses, followed by a subscript that indicates the number of such groups. For example, the condensed structural formula for methanol is CH3OH, which tells us that the molecule contains a CH3 unit that looks like a fragment of methane (CH4). Methanol can therefore be viewed either as a methane molecule in which one hydrogen atom has been replaced by an –OH group or as a water molecule in which one hydrogen atom has been replaced by a –CH3 fragment. Because of their ease of use and information content, we use condensed structural formulas for molecules throughout this text. Ball-and-stick models are used when needed to illustrate the three-dimensional structure of molecules, and space-filling models are used only when it is necessary to visualize the relative sizes of atoms or molecules to understand an important point.
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EXAMPLE 2 Write the molecular formula for each compound. The condensed structural formula is given. a. Sulfur monochloride (also called disulfur dichloride) is a vile-smelling, corrosive yellow liquid used in the production of synthetic rubber. Its condensed structural formula is ClSSCl. b. Ethylene glycol is the major ingredient in antifreeze. Its condensed structural formula is HOCH2CH2OH. c. Trimethylamine is one of the substances responsible for the smell of spoiled fish. Its condensed structural formula is (CH 3)3N. Given: condensed structural formula Asked for: molecular formula Strategy: A Identify every element in the condensed structural formula and then determine whether the compound is organic or inorganic. B As appropriate, use either organic or inorganic convention to list the elements. Then add appropriate subscripts to indicate the number of atoms of each element present in the molecular formula. Solution: The molecular formula lists the elements in the molecule and the number of atoms of each. a. A Each molecule of sulfur monochloride has two sulfur atoms and two chlorine atoms. Because it does not contain mostly carbon and hydrogen, it is an inorganic compound. B Sulfur lies to the left of chlorine in the periodic table, so it is written first in the formula. Adding subscripts gives the molecular formula S2Cl2. b. A Counting the atoms in ethylene glycol, we get six hydrogen atoms, two carbon atoms, and two oxygen atoms per molecule. The compound consists mostly of carbon and hydrogen atoms, so it is organic. B As with all organic compounds, C and H are written first in the molecular
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formula. Adding appropriate subscripts gives the molecular formula C2H 6O2. c. A The condensed structural formula shows that trimethylamine contains three CH3 units, so we have one nitrogen atom, three carbon atoms, and nine hydrogen atoms per molecule. Because trimethylamine contains mostly carbon and hydrogen, it is an organic compound. B According to the convention for organic compounds, C and H are written first, giving the molecular formula C3H9N.
Exercise Write the molecular formula for each molecule. a. Chloroform, which was one of the first anesthetics and was used in many cough syrups until recently, contains one carbon atom, one hydrogen atom, and three chlorine atoms. Its condensed structural formula is CHCl3. b. Hydrazine is used as a propellant in the attitude jets of the space shuttle. Its condensed structural formula is H2NNH2. c. Putrescine is a pungent-smelling compound first isolated from extracts of rotting meat. Its condensed structural formula is H2NCH2CH2CH2CH2NH2. This is often written as H2N(CH2)4NH2 to indicate that there are four CH2 fragments linked together.
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Answer: a. CHCl3 b. N2H4 c. C4H12N2
Ionic Compounds The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively charged protons in the nucleus is equal to the number of negatively charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations17. Conversely, ions that contain more electrons than protons have a net negative charge and are called anions18. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge.
Note the Pattern Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.
17. An ion that has fewer electrons than protons, resulting in a net positive charge. 18. An ion that has fewer protons than electrons, resulting in a net negative charge.
2.1 Chemical Compounds
In covalent compounds, electrons are shared between bonded atoms and are simultaneously attracted to more than one nucleus. In contrast, ionic compounds contain cations and anions rather than discrete neutral molecules. Ionic compounds are held together by the attractive electrostatic interactions between cations and anions. In an ionic compound, the cations and anions are arranged in space to form an extended three-dimensional array that maximizes the number of attractive electrostatic interactions and minimizes the number of repulsive electrostatic interactions (Figure 2.5 "Covalent and Ionic Bonding"). As shown in Equation 2.1, the electrostatic energy of the interaction between two charged particles is proportional to the product of the charges on the particles and inversely proportional to the distance between them: Equation 2.1
electrostatic energy ∝
Q1 Q2 r 136
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where Q1 and Q2 are the electrical charges on particles 1 and 2, and r is the distance between them. When Q1 and Q2 are both positive, corresponding to the charges on cations, the cations repel each other and the electrostatic energy is positive. When Q1 and Q2 are both negative, corresponding to the charges on anions, the anions repel each other and the electrostatic energy is again positive. The electrostatic energy is negative only when the charges have opposite signs; that is, positively charged species are attracted to negatively charged species and vice versa. As shown in Figure 2.6 "The Effect of Charge and Distance on the Strength of Electrostatic Interactions", the strength of the interaction is proportional to the magnitude of the charges and decreases as the distance between the particles increases. We will return to these energetic factors in Chapter 8 "Ionic versus Covalent Bonding", where they are described in greater quantitative detail.
Note the Pattern If the electrostatic energy is positive, the particles repel each other; if the electrostatic energy is negative, the particles are attracted to each other.
Figure 2.5 Covalent and Ionic Bonding
(a) In molecular hydrogen (H2), two hydrogen atoms share two electrons to form a covalent bond. (b) The ionic compound NaCl forms when electrons from sodium atoms are transferred to chlorine atoms. The resulting Na + and Cl− ions form a three-dimensional solid that is held together by attractive electrostatic interactions.
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Figure 2.6 The Effect of Charge and Distance on the Strength of Electrostatic Interactions
As the charge on ions increases or the distance between ions decreases, so does the strength of the attractive (−…+) or repulsive (−…− or +…+) interactions. The strength of these interactions is represented by the thickness of the arrows.
One example of an ionic compound is sodium chloride (NaCl; Figure 2.7 "Sodium Chloride: an Ionic Solid"), formed from sodium and chlorine. In forming chemical compounds, many elements have a tendency to gain or lose enough electrons to attain the same number of electrons as the noble gas closest to them in the periodic table. When sodium and chlorine come into contact, each sodium atom gives up an electron to become a Na+ ion, with 11 protons in its nucleus but only 10 electrons (like neon), and each chlorine atom gains an electron to become a Cl − ion, with 17 protons in its nucleus and 18 electrons (like argon), as shown in part (b) in Figure 2.5 "Covalent and Ionic Bonding". Solid sodium chloride contains equal numbers of cations (Na+) and anions (Cl−), thus maintaining electrical neutrality. Each Na+ ion is surrounded by 6 Cl− ions, and each Cl− ion is surrounded by 6 Na+ ions. Because of the large number of attractive Na+Cl− interactions, the total attractive electrostatic energy in NaCl is great.
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Figure 2.7 Sodium Chloride: an Ionic Solid
The planes of an NaCl crystal reflect the regular three-dimensional arrangement of its Na + (purple) and Cl− (green) ions.
19. An ion with only a single atom.
2.1 Chemical Compounds
Consistent with a tendency to have the same number of electrons as the nearest noble gas, when forming ions, elements in groups 1, 2, and 3 tend to lose one, two, and three electrons, respectively, to form cations, such as Na + and Mg2+. They then have the same number of electrons as the nearest noble gas: neon. Similarly, K +, Ca2+, and Sc3+ have 18 electrons each, like the nearest noble gas: argon. In addition, the elements in group 13 lose three electrons to form cations, such as Al3+, again attaining the same number of electrons as the noble gas closest to them in the periodic table. Because the lanthanides and actinides formally belong to group 3, the most common ion formed by these elements is M3+, where M represents the metal. Conversely, elements in groups 17, 16, and 15 often react to gain one, two, and three electrons, respectively, to form ions such as Cl −, S2−, and P3−. Ions such as these, which contain only a single atom, are called monatomic ions19. You can predict the charges of most monatomic ions derived from the main group elements by simply looking at the periodic table and counting how many columns an element lies from the extreme left or right. For example, you can predict that barium (in group 2) will form Ba2+ to have the same number of electrons as its nearest noble
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gas, xenon, that oxygen (in group 16) will form O2− to have the same number of electrons as neon, and cesium (in group 1) will form Cs + to also have the same number of electrons as xenon. Note that this method does not usually work for most of the transition metals, as you will learn in Section 2.3 "Naming Ionic Compounds". Some common monatomic ions are in Table 2.2 "Some Common Monatomic Ions and Their Names".
Note the Pattern Elements in groups 1, 2, and 3 tend to form 1+, 2+, and 3+ ions, respectively; elements in groups 15, 16, and 17 tend to form 3−, 2−, and 1− ions, respectively.
Table 2.2 Some Common Monatomic Ions and Their Names Group 1
Group 2
Group 3
Group 13
Group 15
Group 16 Group 17
N3− Li+
Be2+
O2−
F−
oxide
fluoride
S2−
Cl−
nitride lithium
beryllium (azide)
Na+
Mg2+
Al3+
sodium
magnesium
aluminum phosphide sulfide
chloride
K+
Ca2+
Sc3+
Ga3+
As3−
Se2−
Br−
scandium
gallium
arsenide
selenide bromide
potassium calcium
2.1 Chemical Compounds
P3−
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Group 1
2.1 Chemical Compounds
Group 2
Group 3
Group 13
Group 15
Group 16 Group 17
Rb+
Sr2+
Y3+
In3+
Te2−
I−
rubidium
strontium
yttrium
indium
telluride iodide
Cs+
Ba2+
La3+
cesium
barium
lanthanum
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EXAMPLE 3 Predict the charge on the most common monatomic ion formed by each element. a. aluminum, used in the quantum logic clock, the world’s most precise clock b. selenium, used to make ruby-colored glass c. yttrium, used to make high-performance spark plugs Given: element Asked for: ionic charge Strategy: A Identify the group in the periodic table to which the element belongs. Based on its location in the periodic table, decide whether the element is a metal, which tends to lose electrons; a nonmetal, which tends to gain electrons; or a semimetal, which can do either. B After locating the noble gas that is closest to the element, determine the number of electrons the element must gain or lose to have the same number of electrons as the nearest noble gas. Solution: a. A Aluminum is a metal in group 13; consequently, it will tend to lose electrons. B The nearest noble gas to aluminum is neon. Aluminum will lose three electrons to form the Al3+ ion, which has the same number of electrons as neon. b. A Selenium is a nonmetal in group 16, so it will tend to gain electrons. B The nearest noble gas is krypton, so we predict that selenium will gain two electrons to form the Se2− ion, which has the same number of electrons as krypton. c. A Yttrium is in group 3, and elements in this group are metals that tend to lose electrons. B The nearest noble gas to yttrium is krypton, so yttrium is predicted to lose three electrons to form Y3+, which has the same number of electrons as krypton. Exercise
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Predict the charge on the most common monatomic ion formed by each element. a. calcium, used to prevent osteoporosis b. iodine, required for the synthesis of thyroid hormones c. zirconium, widely used in nuclear reactors Answer: a. Ca2+ b. I− c. Zr4+
Physical Properties of Ionic and Covalent Compounds In general, ionic and covalent compounds have different physical properties. Ionic compounds usually form hard crystalline solids that melt at rather high temperatures and are very resistant to evaporation. These properties stem from the characteristic internal structure of an ionic solid, illustrated schematically in part (a) in Figure 2.8 "Interactions in Ionic and Covalent Solids", which shows the threedimensional array of alternating positive and negative ions held together by strong electrostatic attractions. In contrast, as shown in part (b) in Figure 2.8 "Interactions in Ionic and Covalent Solids", most covalent compounds consist of discrete molecules held together by comparatively weak intermolecular forces (the forces between molecules), even though the atoms within each molecule are held together by strong intramolecular covalent bonds (the forces within the molecule). Covalent substances can be gases, liquids, or solids at room temperature and pressure, depending on the strength of the intermolecular interactions. Covalent molecular solids tend to form soft crystals that melt at rather low temperatures and evaporate relatively easily.Some covalent substances, however, are not molecular but consist of infinite three-dimensional arrays of covalently bonded atoms and include some of the hardest materials known, such as diamond. This topic will be addressed in Chapter 12 "Solids". The covalent bonds that hold the atoms together in the molecules are unaffected when covalent substances melt or evaporate, so a liquid or vapor of discrete, independent molecules is formed. For example, at room temperature, methane, the major constituent of natural gas, is a gas that is composed of discrete CH4 molecules. A comparison of the different physical properties of ionic compounds and covalent molecular substances is given in Table 2.3 "The Physical Properties of Typical Ionic Compounds and Covalent Molecular Substances".
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Table 2.3 The Physical Properties of Typical Ionic Compounds and Covalent Molecular Substances Ionic Compounds
Covalent Molecular Substances
hard solids
gases, liquids, or soft solids
high melting points low melting points nonvolatile
volatile
Figure 2.8 Interactions in Ionic and Covalent Solids
(a) The positively and negatively charged ions in an ionic solid such as sodium chloride (NaCl) are held together by strong electrostatic interactions. (b) In this representation of the packing of methane (CH 4) molecules in solid methane, a prototypical molecular solid, the methane molecules are held together in the solid only by relatively weak intermolecular forces, even though the atoms within each methane molecule are held together by strong covalent bonds.
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Summary The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure.
KEY TAKEAWAY • There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties.
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CONCEPTUAL PROBLEMS 1. Ionic and covalent compounds are held together by electrostatic attractions between oppositely charged particles. Describe the differences in the nature of the attractions in ionic and covalent compounds. Which class of compounds contains pairs of electrons shared between bonded atoms? 2. Which contains fewer electrons than the neutral atom—the corresponding cation or the anion? 3. What is the difference between an organic compound and an inorganic compound? 4. What is the advantage of writing a structural formula as a condensed formula? 5. The majority of elements that exist as diatomic molecules are found in one group of the periodic table. Identify the group. 6. Discuss the differences between covalent and ionic compounds with regard to a. the forces that hold the atoms together. b. melting points. c. physical states at room temperature and pressure. 7. Why do covalent compounds generally tend to have lower melting points than ionic compounds?
ANSWER 7. Covalent compounds generally melt at lower temperatures than ionic compounds because the intermolecular interactions that hold the molecules together in a molecular solid are weaker than the electrostatic attractions that hold oppositely charged ions together in an ionic solid.
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NUMERICAL PROBLEMS 1. The structural formula for chloroform (CHCl3) was shown in Example 2. Based on this information, draw the structural formula of dichloromethane (CH 2Cl2). 2. What is the total number of electrons present in each ion? a. b. c. d. e. f. g.
F− Rb+ Ce3+ Zr4+ Zn2+ Kr2+ B3+
3. What is the total number of electrons present in each ion? a. b. c. d. e. f. g.
Ca2+ Se2− In3+ Sr2+ As3+ N3− Tl+
4. Predict how many electrons are in each ion. a. b. c. d. e. f.
an oxygen ion with a −2 charge a beryllium ion with a +2 charge a silver ion with a +1 charge a selenium ion with a +4 charge an iron ion with a +2 charge a chlorine ion with a −1 charge
5. Predict how many electrons are in each ion. a. b. c. d. e. f.
a copper ion with a +2 charge a molybdenum ion with a +4 charge an iodine ion with a −1 charge a gallium ion with a +3 charge an ytterbium ion with a +3 charge a scandium ion with a +3 charge
6. Predict the charge on the most common monatomic ion formed by each element.
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a. b. c. d. e. f.
chlorine phosphorus scandium magnesium arsenic oxygen
7. Predict the charge on the most common monatomic ion formed by each element. a. b. c. d. e. f.
sodium selenium barium rubidium nitrogen aluminum
8. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. 9
a. 4 X 2+ 1 b. 1 X – 16
c. 8
X 2–
9. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. 7
a. 3 X + 19 b. 9 X – 27
c. 13 X 3+
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ANSWERS 5. a. b. c. d. e. f.
27 38 54 28 67 18
9. a. Li, Li+, 2nd period, group 1 b. F, F–, 2nd period, group 17 c. Al, Al3+, 3nd period, group 13
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2.2 Chemical Formulas LEARNING OBJECTIVE 1. To describe the composition of a chemical compound.
When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula20, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is C4H10. The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore C 2H5. The formula unit21 is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane, for example, has the empirical formula C2H5, but it contains two C2H5 formula units, giving a molecular formula of C4H10.
20. A formula for a compound that consists of the atomic symbol for each component element accompanied by a subscript indicating the relative number of atoms of that element in the compound, reduced to the smallest whole numbers.
Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as Na+ and Cl−, then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be NaCl. If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of Mg2+ and Cl−, for example, two Cl− ions are needed to balance the two positive charges on each Mg 2+ ion, giving an empirical formula of MgCl2. Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O.
21. The absolute grouping of atoms or ions represented by the empirical formula.
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Note the Pattern Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions.
Binary Ionic Compounds An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound22. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows:
Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges.
When crossing charges, you will sometimes find it necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula Mg 2O2:
This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion.
22. An ionic compound that contains only two elements, one present as a cation and one as an anion.
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EXAMPLE 4 Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. a. Ga3+ and As3− b. Eu3+ and O2− c. calcium and chlorine Given: ions or elements Asked for: empirical formula for binary ionic compound Strategy: A If not given, determine the ionic charges based on the location of the elements in the periodic table. B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers to write the empirical formula. Check to make sure the empirical formula is electrically neutral. Solution:
a. B Using the absolute values of the charges on the ions as the subscripts gives Ga3As3:
Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga3+ and As3− have charges of equal magnitude but opposite signs. One Ga3+ ion balances the charge on one As3− ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is
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gallium arsenide, which is widely used in the electronics industry in transistors and other devices. b. B Because Eu3+ has a charge of +3 and O2− has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion:
The subscript for Eu3+ is 2 (from O2−), and the subscript for O2− is 3 (from Eu3+), giving Eu2O3; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu2O3 is neutral. Europium oxide is responsible for the red color in television and computer screens. c. A Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca2+. Chlorine lies in group 17, so it should gain one electron to form Cl−. B Two Cl− ions are needed to balance the charge on one Ca2+ ion, which leads to the empirical formula CaCl2. We could also cross charges, using the absolute value of the charge on Ca2+ as the subscript for Cl and the absolute value of the charge on Cl − as the subscript for Ca:
The subscripts in CaCl2 cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter. Exercise
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Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair. a. Li+ and N3− b. Al3+ and O2− c. lithium and oxygen Answer: a. Li3N b. Al2O3 c. Li2O
Polyatomic Ions Polyatomic ions23 are groups of atoms that bear a net electrical charge, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH 4+) and the methylammonium (CH3NH3+) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in Table 2.4 "Common Polyatomic Ions and Their Names". Table 2.4 Common Polyatomic Ions and Their Names Formula
23. A group of two or more atoms that has a net electrical charge.
2.2 Chemical Formulas
Name of Ion
NH4+
ammonium
CH3NH3+
methylammonium
OH−
hydroxide
O22−
peroxide
CN−
cyanide
SCN−
thiocyanate
NO2−
nitrite
NO3−
nitrate
CO32−
carbonate
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Formula
Name of Ion
HCO3−
hydrogen carbonate, or bicarbonate
SO32−
sulfite
SO42−
sulfate
HSO4−
hydrogen sulfate, or bisulfate
PO43−
phosphate
HPO42−
hydrogen phosphate
H2PO4−
dihydrogen phosphate
ClO−
hypochlorite
ClO2−
chlorite
ClO3−
chlorate
ClO4−
perchlorate
MnO4−
permanganate
CrO42−
chromate
Cr2O72−
dichromate
C2O42−
oxalate
HCO2−
formate
CH3CO2−
acetate
C6H5CO2− benzoate
The method we used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus K+ and NO3− ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42− form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3− form NH4NO3 (ammonium nitrate), which is a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43− ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca 2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca 3(PO4)2; the
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parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43− ions.
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EXAMPLE 5 Write the empirical formula for the compound formed from each ion pair. a. Na+ and HPO42− b. potassium cation and cyanide anion c. calcium cation and hypochlorite anion Given: ions Asked for: empirical formula for ionic compound Strategy: A If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use Table 2.4 "Common Polyatomic Ions and Their Names" to find the charge on a polyatomic ion. B Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula. Solution: a. B Because HPO42− has a charge of −2 and Na+ has a charge of +1, the empirical formula requires two Na+ ions to balance the charge of the polyatomic ion, giving Na2HPO4. The subscripts are reduced to the lowest numbers, so the empirical formula is Na2HPO4. This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts. b. A The potassium cation is K+, and the cyanide anion is CN−. B Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally. c. A The calcium cation is Ca2+, and the hypochlorite anion is ClO−. B Two ClO− ions are needed to balance the charge on one Ca2+ ion, giving Ca(ClO)2. The subscripts cannot be reduced further, so the empirical formula is Ca(ClO)2. This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools.
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Exercise Write the empirical formula for the compound formed from each ion pair. a. Ca2+ and H2PO4− b. sodium cation and bicarbonate anion c. ammonium cation and sulfate anion Answer: a. Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder. b. NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda. c. (NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers.
Hydrates Many ionic compounds occur as hydrates24, compounds that contain specific ratios of loosely bound water molecules, called waters of hydration25. Waters of hydration can often be removed simply by heating. For example, calcium dihydrogen phosphate can form a solid that contains one molecule of water per Ca(H2PO4)2 unit and is used as a leavening agent in the food industry to cause baked goods to rise. The empirical formula for the solid is Ca(H 2PO4)2·H2O. In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula CuSO4·5H2O. When heated, all five water molecules are lost, giving a white solid with the empirical formula CuSO 4 (Figure 2.9 "Loss of Water from a Hydrate with Heating").
24. A compound that contains specific ratios of loosely bound water molecules, called waters of hydration. 25. The loosely bound water molecules in hydrate compounds. These waters of hydration can often be removed by simply heating the compound.
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Figure 2.9 Loss of Water from a Hydrate with Heating
When blue CuSO4·5H2O is heated, two molecules of water are lost at 30°C, two more at 110°C, and the last at 250°C to give white CuSO4.
Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, CaSO4·½H2O is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas CaSO 4·2H2O is the less dense, flakier gypsum, a mineral used in drywall panels for home construction. When a cast would set, a mixture of plaster of Paris and water crystallized to give solid CaSO4·2H2O. Similar processes are used in the setting of cement and concrete.
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Summary An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration.
KEY TAKEAWAY • The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.
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CONCEPTUAL PROBLEMS 1. What are the differences and similarities between a polyatomic ion and a molecule? 2. Classify each compound as ionic or covalent. a. b. c. d. e. f.
Zn3(PO4)2 C6H5CO2H K2Cr2O7 CH3CH2SH NH4Br CCl2F2
3. Classify each compound as ionic or covalent. Which are organic compounds and which are inorganic compounds? a. b. c. d. e.
CH3CH2CO2H CaCl2 Y(NO3)3 H2S NaC2H3O2
4. Generally, one cannot determine the molecular formula directly from an empirical formula. What other information is needed? 5. Give two pieces of information that we obtain from a structural formula that we cannot obtain from an empirical formula. 6. The formulas of alcohols are often written as ROH rather than as empirical formulas. For example, methanol is generally written as CH 3OH rather than CH4O. Explain why the ROH notation is preferred. 7. The compound dimethyl sulfide has the empirical formula C 2H6S and the structural formula CH3SCH3. What information do we obtain from the structural formula that we do not get from the empirical formula? Write the condensed structural formula for the compound. 8. What is the correct formula for magnesium hydroxide—MgOH2 or Mg(OH)2? Why? 9. Magnesium cyanide is written as Mg(CN)2, not MgCN2. Why? 10. Does a given hydrate always contain the same number of waters of hydration?
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ANSWER 7. The structural formula gives us the connectivity of the atoms in the molecule or ion, as well as a schematic representation of their arrangement in space. Empirical formulas tell us only the ratios of the atoms present. The condensed structural formula of dimethylsulfide is (CH3)2S.
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NUMERICAL PROBLEMS 1. Write the formula for each compound. a. magnesium sulfate, which has 1 magnesium atom, 4 oxygen atoms, and 1 sulfur atom b. ethylene glycol (antifreeze), which has 6 hydrogen atoms, 2 carbon atoms, and 2 oxygen atoms c. acetic acid, which has 2 oxygen atoms, 2 carbon atoms, and 4 hydrogen atoms d. potassium chlorate, which has 1 chlorine atom, 1 potassium atom, and 3 oxygen atoms e. sodium hypochlorite pentahydrate, which has 1 chlorine atom, 1 sodium atom, 6 oxygen atoms, and 10 hydrogen atoms 2. Write the formula for each compound. a. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms b. barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms c. iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms d. manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms e. silver phosphate, which has 1 phosphorus atom, 3 silver atoms, and 4 oxygen atoms 3. Complete the following table by filling in the formula for the ionic compound formed by each cation-anion pair. Ion Cl−
+ K+ Fe3+ NH4 Ba2+ KCl
SO42− PO43− NO3− OH− 4. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. a. zinc and sulfur
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b. c. d. e.
barium and iodine magnesium and chlorine silicon and oxygen sodium and sulfur
5. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. a. b. c. d. e.
lithium and nitrogen cesium and chlorine germanium and oxygen rubidium and sulfur arsenic and sodium
6. Write the empirical formula for each compound. a. b. c. d. e.
Na2S2O4 B 2H 6 C6H12O6 P4O10 KMnO4
7. Write the empirical formula for each compound. a. b. c. d. e.
2.2 Chemical Formulas
Al2Cl6 K2Cr2O7 C2H4 (NH2)2CNH CH3COOH
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ANSWERS 1. a. b. c. d. e.
MgSO4 C2H 6O2 C2H 4O2 KClO3 NaOCl·5H2O 3.
Ion Cl −
K+ KCl
Fe 3+ FeCl3
NH 4 + NH4Cl
Ba 2+ BaCl2
SO 4 2− K2SO4 Fe2(SO4)3 (NH4)2SO4 BaSO4
2.2 Chemical Formulas
PO 4 3− K3PO4 FePO4
(NH4)3PO4 Ba3(PO4)2
NO 3 −
KNO3
Fe(NO3)3
NH4NO3
Ba(NO3)2
OH −
KOH
Fe(OH)3
NH4OH
Ba(OH)2
5. a. b. c. d. e.
Li3N CsCl GeO2 Rb2S Na3As
7. a. b. c. d. e.
AlCl3 K2Cr2O7 CH2 CH5N3 CH2O
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2.3 Naming Ionic Compounds LEARNING OBJECTIVE 1. To name ionic compounds.
The empirical and molecular formulas discussed in the preceding section are precise and highly informative, but they have some disadvantages. First, they are inconvenient for routine verbal communication. For example, saying “C-A-three-PO-four-two” for Ca3(PO4)2 is much more difficult than saying “calcium phosphate.” In addition, you will see in Section 2.4 "Naming Covalent Compounds" that many compounds have the same empirical and molecular formulas but different arrangements of atoms, which result in very different chemical and physical properties. In such cases, it is necessary for the compounds to have different names that distinguish among the possible arrangements. Many compounds, particularly those that have been known for a relatively long time, have more than one name: a common name (sometimes more than one) and a systematic name, which is the name assigned by adhering to specific rules. Like the names of most elements, the common names of chemical compounds generally have historical origins, although they often appear to be unrelated to the compounds of interest. For example, the systematic name for KNO 3 is potassium nitrate, but its common name is saltpeter. In this text, we use a systematic nomenclature to assign meaningful names to the millions of known substances. Unfortunately, some chemicals that are widely used in commerce and industry are still known almost exclusively by their common names; in such cases, you must be familiar with the common name as well as the systematic one. The objective of this and the next two sections is to teach you to write the formula for a simple inorganic compound from its name—and vice versa—and introduce you to some of the more frequently encountered common names. We begin with binary ionic compounds, which contain only two elements. The procedure for naming such compounds is outlined in Figure 2.10 "Naming an Ionic Compound" and uses the following steps:
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Figure 2.10 Naming an Ionic Compound
1. Place the ions in their proper order: cation and then anion. 2. Name the cation. a. Metals that form only one cation. As noted in Section 2.1 "Chemical Compounds", these metals are usually in groups 1–3, 12, and 13. The name of the cation of a metal that forms only one cation is the same as the name of the metal (with the word ion added if the cation is by itself). For example, Na + is the sodium ion, Ca2+ is the calcium ion, and Al3+ is the aluminum ion. b. Metals that form more than one cation. As shown in Figure 2.11 "Metals That Form More Than One Cation and Their Locations in the Periodic Table", many metals can form more than one cation. This behavior is observed for most transition metals, many actinides, and the heaviest elements of groups 13–15. In such cases, the positive charge on the metal is indicated by a roman numeral in parentheses immediately following the name of the metal. Thus Cu+ is copper(I) (read as “copper one”), Fe2+ is iron(II), Fe3+ is iron(III), Sn2+ is tin(II), and Sn4+ is tin(IV). An older system of nomenclature for such cations is still widely used, however. The name of the cation with the higher charge is formed from the root of the element’s Latin name with the suffix -ic attached, and the name of the cation with the lower charge has the same root with the suffix -ous. The names of Fe3+, Fe2+, Sn4+, and Sn2+ are therefore ferric, ferrous, stannic, and stannous, respectively. Even though this text uses the systematic names with roman numerals, you should be able to recognize these common names because they are still often used. For example, on the label
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of your dentist’s fluoride rinse, the compound chemists call tin(II) fluoride is usually listed as stannous fluoride. Some examples of metals that form more than one cation are in Table 2.5 "Common Cations of Metals That Form More Than One Ion" along with the names of the ions. Note that the simple Hg + cation does not occur in chemical compounds. Instead, all compounds of mercury(I) contain a dimeric cation, Hg22+, in which the two Hg atoms are bonded together. Table 2.5 Common Cations of Metals That Form More Than One Ion Cation
Systematic Name
Common Name
Cation
Systematic Name
Common Name
Cr2+
chromium(II)
chromous
Cu2+
copper(II)
cupric
Cr3+
chromium(III)
chromic
Cu+
copper(I)
cuprous
Mn2+
manganese(II)
manganous* Hg2+
Mn3+
manganese(III) manganic*
Hg22+
mercury(I)
mercurous†
Fe2+
iron(II)
ferrous
Sn4+
tin(IV)
stannic
Fe3+
iron(III)
ferric
Sn2+
tin(II)
stannous
Co2+
cobalt(II)
cobaltous*
Pb4+
lead(IV)
plumbic*
Co3+
cobalt(III)
cobaltic*
Pb2+
lead(II)
plumbous*
mercury(II) mercuric
* Not widely used. †
The isolated mercury(I) ion exists only as the gaseous ion.
c. Polyatomic cations. The names of the common polyatomic cations that are relatively important in ionic compounds (such as, the ammonium ion) are in Table 2.4 "Common Polyatomic Ions and Their Names". 3. Name the anion. a. Monatomic anions. Monatomic anions are named by adding the suffix -ide to the root of the name of the parent element; thus, Cl− is chloride, O2− is oxide, P3− is phosphide, N3− is nitride (also called azide), and C4− is carbide. Because the charges on these ions can be predicted from their position in the periodic table, it is not necessary to specify the charge in the name. Examples of
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monatomic anions are in Table 2.2 "Some Common Monatomic Ions and Their Names". b. Polyatomic anions. Polyatomic anions typically have common names that you must learn; some examples are in Table 2.4 "Common Polyatomic Ions and Their Names". Polyatomic anions that contain a single metal or nonmetal atom plus one or more oxygen atoms are called oxoanions (or oxyanions). In cases where only two oxoanions are known for an element, the name of the oxoanion with more oxygen atoms ends in -ate, and the name of the oxoanion with fewer oxygen atoms ends in -ite. For example, NO3− is nitrate and NO2− is nitrite. The halogens and some of the transition metals form more extensive series of oxoanions with as many as four members. In the names of these oxoanions, the prefix per- is used to identify the oxoanion with the most oxygen (so that ClO4− is perchlorate and ClO3− is chlorate), and the prefix hypo- is used to identify the anion with the fewest oxygen (ClO2− is chlorite and ClO− is hypochlorite). The relationship between the names of oxoanions and the number of oxygen atoms present is diagrammed in Figure 2.12 "The Relationship between the Names of Oxoanions and the Number of Oxygen Atoms Present". Differentiating the oxoanions in such a series is no trivial matter. For example, the hypochlorite ion is the active ingredient in laundry bleach and swimming pool disinfectant, but compounds that contain the perchlorate ion can explode if they come into contact with organic substances. 4. Write the name of the compound as the name of the cation followed by the name of the anion. It is not necessary to indicate the number of cations or anions present per formula unit in the name of an ionic compound because this information is implied by the charges on the ions. You must consider the charge of the ions when writing the formula for an ionic compound from its name, however. Because the charge on the chloride ion is −1 and the charge on the calcium ion is +2, for example, consistent with their positions in the periodic table, simple arithmetic tells you that calcium chloride must contain twice as many chloride ions as calcium ions to maintain electrical neutrality. Thus the formula is CaCl 2. Similarly, calcium phosphate must be Ca3(PO4)2 because the cation and the anion have charges of +2 and −3, respectively. The best way to learn how to name ionic compounds is to work through a few examples, referring to Figure 2.10 "Naming an Ionic Compound", Table 2.2 "Some
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Common Monatomic Ions and Their Names", Table 2.4 "Common Polyatomic Ions and Their Names", and Table 2.5 "Common Cations of Metals That Form More Than One Ion" as needed.
Figure 2.11 Metals That Form More Than One Cation and Their Locations in the Periodic Table
With only a few exceptions, these metals are usually transition metals or actinides.
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Figure 2.12 The Relationship between the Names of Oxoanions and the Number of Oxygen Atoms Present
Note the Pattern Cations are always named before anions. Most transition metals, many actinides, and the heaviest elements of groups 13–15 can form more than one cation.
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EXAMPLE 6 Write the systematic name (and the common name if applicable) for each ionic compound. a. b. c. d.
LiCl MgSO4 (NH4)3PO4 Cu2O
Given: empirical formula Asked for: name Strategy: A If only one charge is possible for the cation, give its name, consulting Table 2.2 "Some Common Monatomic Ions and Their Names" or Table 2.4 "Common Polyatomic Ions and Their Names" if necessary. If the cation can have more than one charge (Table 2.5 "Common Cations of Metals That Form More Than One Ion"), specify the charge using roman numerals. B If the anion does not contain oxygen, name it according to step 3a, using Table 2.2 "Some Common Monatomic Ions and Their Names" and Table 2.4 "Common Polyatomic Ions and Their Names" if necessary. For polyatomic anions that contain oxygen, use Table 2.4 "Common Polyatomic Ions and Their Names" and the appropriate prefix and suffix listed in step 3b. C Beginning with the cation, write the name of the compound. Solution: a. A B Lithium is in group 1, so we know that it forms only the Li + cation, which is the lithium ion. Similarly, chlorine is in group 7, so it forms the Cl− anion, which is the chloride ion. C Because we begin with the name of the cation, the name of this compound is lithium chloride, which is used medically as an antidepressant drug. b. A B The cation is the magnesium ion, and the anion, which contains oxygen, is sulfate. C Because we list the cation first, the name of this compound is magnesium sulfate. A hydrated form of magnesium sulfate
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(MgSO4·7H2O) is sold in drugstores as Epsom salts, a harsh but effective laxative. c. A B The cation is the ammonium ion (from Table 2.4 "Common Polyatomic Ions and Their Names"), and the anion is phosphate. C The compound is therefore ammonium phosphate, which is widely used as a fertilizer. It is not necessary to specify that the formula unit contains three ammonium ions because three are required to balance the negative charge on phosphate. d. A B The cation is a transition metal that often forms more than one cation (Table 2.5 "Common Cations of Metals That Form More Than One Ion"). We must therefore specify the positive charge on the cation in the name: copper(I) or, according to the older system, cuprous. The anion is oxide. C The name of this compound is copper(I) oxide or, in the older system, cuprous oxide. Copper(I) oxide is used as a red glaze on ceramics and in antifouling paints to prevent organisms from growing on the bottoms of boats.
Cu2O. The bottom of a boat is protected with a red antifouling paint containing copper(I) oxide, Cu2O.
Exercise Write the systematic name (and the common name if applicable) for each ionic compound. a. CuCl2
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b. MgCO3 c. FePO4 Answer: a. copper(II) chloride (or cupric chloride) b. magnesium carbonate c. iron(III) phosphate (or ferric phosphate)
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EXAMPLE 7 Write the formula for each compound. a. calcium dihydrogen phosphate b. aluminum sulfate c. chromium(III) oxide Given: systematic name Asked for: formula Strategy: A Identify the cation and its charge using the location of the element in the periodic table and Table 2.2 "Some Common Monatomic Ions and Their Names", Table 2.3 "The Physical Properties of Typical Ionic Compounds and Covalent Molecular Substances", Table 2.4 "Common Polyatomic Ions and Their Names", and Table 2.5 "Common Cations of Metals That Form More Than One Ion". If the cation is derived from a metal that can form cations with different charges, use the appropriate roman numeral or suffix to indicate its charge. B Identify the anion using Table 2.2 "Some Common Monatomic Ions and Their Names" and Table 2.4 "Common Polyatomic Ions and Their Names". Beginning with the cation, write the compound’s formula and then determine the number of cations and anions needed to achieve electrical neutrality. Solution: a. A Calcium is in group 2, so it forms only the Ca2+ ion. B Dihydrogen phosphate is the H2PO4− ion (Table 2.4 "Common Polyatomic Ions and Their Names"). Two H2PO4− ions are needed to balance the positive charge on Ca2+, to give Ca(H2PO4)2. A hydrate of calcium dihydrogen phosphate, Ca(H2PO4)2·H2O, is the active ingredient in baking powder. b. A Aluminum, near the top of group 13 in the periodic table, forms only one cation, Al3+ (Figure 2.11 "Metals That Form More Than One Cation and Their Locations in the Periodic Table"). B Sulfate is SO42− (Table 2.4 "Common Polyatomic Ions and Their Names"). To balance the electrical charges, we need two Al3+ cations and three SO42− anions, giving
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Al2(SO4)3. Aluminum sulfate is used to tan leather and purify drinking water. c. A Because chromium is a transition metal, it can form cations with different charges. The roman numeral tells us that the positive charge in this case is +3, so the cation is Cr3+. B Oxide is O2−. Thus two cations (Cr3+) and three anions (O2−) are required to give an electrically neutral compound, Cr2O3. This compound is a common green pigment that has many uses, including camouflage coatings.
Cr2O3. Chromium(III) oxide (Cr2O3) is a common pigment in dark green paints, such as camouflage paint.
Exercise Write the formula for each compound. a. barium chloride b. sodium carbonate c. iron(III) hydroxide Answer: a. BaCl2 b. Na2CO3 c. Fe(OH)3
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Summary Ionic compounds are named according to systematic procedures, although common names are widely used. Systematic nomenclature enables us to write the structure of any compound from its name and vice versa. Ionic compounds are named by writing the cation first, followed by the anion. If a metal can form cations with more than one charge, the charge is indicated by roman numerals in parentheses following the name of the metal. Oxoanions are polyatomic anions that contain a single metal or nonmetal atom and one or more oxygen atoms.
KEY TAKEAWAY • There is a systematic method used to name ionic compounds.
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CONCEPTUAL PROBLEMS 1. Name each cation. a. b. c. d. e.
K+ Al3+ NH4+ Mg2+ Li+
2. Name each anion. a. b. c. d. e. f. g. h.
Br− CO32− S2− NO3− HCO2− F− ClO− C2O42−
3. Name each anion. a. b. c. d. e. f. g. h.
PO43− Cl− SO32− CH3CO2− HSO4− ClO4− NO2− O2−
4. Name each anion. a. b. c. d. e. f. g.
SO42− CN− Cr2O72− N3− OH− I− O22−
5. Name each compound.
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a. b. c. d. e. f. g.
MgBr2 NH4CN CaO KClO3 K3PO4 NH4NO2 NaN3
6. Name each compound. a. b. c. d. e. f. g.
NaNO3 Cu3(PO4)2 NaOH Li4C CaF2 NH4Br MgCO3
7. Name each compound. a. b. c. d. e. f. g.
RbBr Mn2(SO4)3 NaClO (NH4)2SO4 NaBr KIO3 Na2CrO4
8. Name each compound. a. b. c. d. e. f. g.
NH4ClO4 SnCl4 Fe(OH)2 Na2O MgCl2 K2SO4 RaCl2
9. Name each compound. a. KCN b. LiOH c. CaCl2
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d. e. f. g.
NiSO4 NH4ClO2 LiClO4 La(CN)3
ANSWER 7. a. b. c. d. e. f. g.
2.3 Naming Ionic Compounds
rubidium bromide manganese(III) sulfate sodium hypochlorite ammonium sulfate sodium bromide potassium iodate sodium chromate
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NUMERICAL PROBLEMS 1. For each ionic compound, name the cation and the anion and give the charge on each ion. a. b. c. d. e. f. g.
BeO Pb(OH)2 BaS Na2Cr2O7 ZnSO4 KClO NaH2PO4
2. For each ionic compound, name the cation and the anion and give the charge on each ion. a. b. c. d. e. f. g.
Zn(NO3)2 CoS BeCO3 Na2SO4 K 2C2O4 NaCN FeCl2
3. Write the formula for each compound. a. b. c. d. e. f. g. h.
magnesium carbonate aluminum sulfate potassium phosphate lead(IV) oxide silicon nitride sodium hypochlorite titanium(IV) chloride disodium ammonium phosphate
4. Write the formula for each compound. a. b. c. d. e. f. g.
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lead(II) nitrate ammonium phosphate silver sulfide barium sulfate cesium iodide sodium bicarbonate potassium dichromate
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h. sodium hypochlorite 5. Write the formula for each compound. a. b. c. d. e.
zinc cyanide silver chromate lead(II) iodide benzene copper(II) perchlorate
6. Write the formula for each compound. a. b. c. d. e.
calcium fluoride sodium nitrate iron(III) oxide copper(II) acetate sodium nitrite
7. Write the formula for each compound. a. b. c. d. e. f. g.
sodium hydroxide calcium cyanide magnesium phosphate sodium sulfate nickel(II) bromide calcium chlorite titanium(IV) bromide
8. Write the formula for each compound. a. b. c. d. e. f. g.
sodium chlorite potassium nitrite sodium nitride (also called sodium azide) calcium phosphide tin(II) chloride calcium hydrogen phosphate iron(II) chloride dihydrate
9. Write the formula for each compound. a. b. c. d. e.
2.3 Naming Ionic Compounds
potassium carbonate chromium(III) sulfite cobalt(II) phosphate magnesium hypochlorite nickel(II) nitrate hexahydrate
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2.4 Naming Covalent Compounds LEARNING OBJECTIVE 1. To name covalent compounds that contain up to three elements.
As with ionic compounds, the system that chemists have devised for naming covalent compounds enables us to write the molecular formula from the name and vice versa. In this and the following section, we describe the rules for naming simple covalent compounds. We begin with inorganic compounds and then turn to simple organic compounds that contain only carbon and hydrogen.
Binary Inorganic Compounds Binary covalent compounds—that is, covalent compounds that contain only two elements—are named using a procedure similar to that used to name simple ionic compounds, but prefixes are added as needed to indicate the number of atoms of each kind. The procedure, diagrammed in Figure 2.13 "Naming a Covalent Inorganic Compound", uses the following steps: 1. Place the elements in their proper order. a. The element farthest to the left in the periodic table is usually named first. If both elements are in the same group, the element closer to the bottom of the column is named first. b. The second element is named as if it were a monatomic anion in an ionic compound (even though it is not), with the suffix -ide attached to the root of the element name.
Figure 2.13 Naming a Covalent Inorganic Compound
2. Identify the number of each type of atom present. a. Prefixes derived from Greek stems are used to indicate the number of each type of atom in the formula unit (Table 2.6 "Prefixes for Indicating the Number of Atoms in Chemical Names"). The prefix
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mono- (“one”) is used only when absolutely necessary to avoid confusion, just as we omit the subscript 1 when writing molecular formulas. To demonstrate steps 1 and 2a, we name HCl as hydrogen chloride (because hydrogen is to the left of chlorine in the periodic table) and PCl5 as phosphorus pentachloride. The order of the elements in the name of BrF3, bromine trifluoride, is determined by the fact that bromine lies below fluorine in group 17. Table 2.6 Prefixes for Indicating the Number of Atoms in Chemical Names Prefix
Number
mono-
1
di-
2
tri-
3
tetra-
4
penta-
5
hexa-
6
hepta-
7
octa-
8
nona-
9
deca-
10
undeca-
11
dodeca-
12
b. If a molecule contains more than one atom of both elements, then prefixes are used for both. Thus N2O3 is dinitrogen trioxide, as shown in Figure 2.13 "Naming a Covalent Inorganic Compound". c. In some names, the final a or o of the prefix is dropped to avoid awkward pronunciation. Thus OsO4 is osmium tetroxide rather than osmium tetraoxide. 3. Write the name of the compound.
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a. Binary compounds of the elements with oxygen are generally named as “element oxide,” with prefixes that indicate the number of atoms of each element per formula unit. For example, CO is carbon monoxide. The only exception is binary compounds of oxygen with fluorine, which are named as oxygen fluorides. (The reasons for this convention will become clear in Chapter 7 "The Periodic Table and Periodic Trends" and Chapter 8 "Ionic versus Covalent Bonding".) b. Certain compounds are always called by the common names that were assigned long ago when names rather than formulas were used. For example, H2O is water (not dihydrogen oxide); NH3 is ammonia; PH3 is phosphine; SiH4 is silane; and B2H6, a dimer of BH3, is diborane. For many compounds, the systematic name and the common name are both used frequently, so you must be familiar with them. For example, the systematic name for NO is nitrogen monoxide, but it is much more commonly called nitric oxide. Similarly, N2O is usually called nitrous oxide rather than dinitrogen monoxide. Notice that the suffixes -ic and -ous are the same ones used for ionic compounds.
Note the Pattern Start with the element at the far left in the periodic table and work to the right. If two or more elements are in the same group, start with the bottom element and work up.
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EXAMPLE 8 Write the name of each binary covalent compound. a. SF6 b. N2O4 c. ClO2 Given: molecular formula Asked for: name of compound Strategy: A List the elements in order according to their positions in the periodic table. Identify the number of each type of atom in the chemical formula and then use Table 2.6 "Prefixes for Indicating the Number of Atoms in Chemical Names" to determine the prefixes needed. B If the compound contains oxygen, follow step 3a. If not, decide whether to use the common name or the systematic name. Solution: a. A Because sulfur is to the left of fluorine in the periodic table, sulfur is named first. Because there is only one sulfur atom in the formula, no prefix is needed. B There are, however, six fluorine atoms, so we use the prefix for six: hexa- (Table 2.6 "Prefixes for Indicating the Number of Atoms in Chemical Names"). The compound is sulfur hexafluoride. b. A Because nitrogen is to the left of oxygen in the periodic table, nitrogen is named first. Because more than one atom of each element is present, prefixes are needed to indicate the number of atoms of each. According to Table 2.6 "Prefixes for Indicating the Number of Atoms in Chemical Names", the prefix for two is di-, and the prefix for four is tetra-. B The compound is dinitrogen tetroxide (omitting the a in tetra- according to step 2c) and is used as a component of some rocket fuels. c. A Although oxygen lies to the left of chlorine in the periodic table, it is not named first because ClO2 is an oxide of an element other than fluorine (step 3a). Consequently, chlorine is named first, but a prefix is not necessary because each molecule has only one atom of chlorine. B Because there are two oxygen atoms, the compound is a dioxide. Thus
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the compound is chlorine dioxide. It is widely used as a substitute for chlorine in municipal water treatment plants because, unlike chlorine, it does not react with organic compounds in water to produce potentially toxic chlorinated compounds. Exercise Write the name of each binary covalent compound. a. IF7 b. N2O5 c. OF2 Answer: a. iodine heptafluoride b. dinitrogen pentoxide c. oxygen difluoride
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EXAMPLE 9 Write the formula for each binary covalent compound. a. sulfur trioxide b. diiodine pentoxide Given: name of compound Asked for: formula Strategy: List the elements in the same order as in the formula, use Table 2.6 "Prefixes for Indicating the Number of Atoms in Chemical Names" to identify the number of each type of atom present, and then indicate this quantity as a subscript to the right of that element when writing the formula. Solution: a. Sulfur has no prefix, which means that each molecule has only one sulfur atom. The prefix tri- indicates that there are three oxygen atoms. The formula is therefore SO3. Sulfur trioxide is produced industrially in huge amounts as an intermediate in the synthesis of sulfuric acid. b. The prefix di- tells you that each molecule has two iodine atoms, and the prefix penta- indicates that there are five oxygen atoms. The formula is thus I2O5, a compound used to remove carbon monoxide from air in respirators. Exercise Write the formula for each binary covalent compound. a. silicon tetrachloride b. disulfur decafluoride Answer: a. SiCl4 b. S2F10
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The structures of some of the compounds in Example 8 and Example 9 are shown in Figure 2.14 "The Structures of Some Covalent Inorganic Compounds and the Locations of the “Central Atoms” in the Periodic Table", along with the location of the “central atom” of each compound in the periodic table. It may seem that the compositions and structures of such compounds are entirely random, but this is not true. After you have mastered the material in Chapter 7 "The Periodic Table and Periodic Trends" and Chapter 8 "Ionic versus Covalent Bonding", you will be able to predict the compositions and structures of compounds of this type with a high degree of accuracy. Figure 2.14 The Structures of Some Covalent Inorganic Compounds and the Locations of the “Central Atoms” in the Periodic Table
The compositions and structures of covalent inorganic compounds are not random. As you will learn in Chapter 7 "The Periodic Table and Periodic Trends" and Chapter 8 "Ionic versus Covalent Bonding", they can be predicted from the locations of the component atoms in the periodic table.
Hydrocarbons Approximately one-third of the compounds produced industrially are organic compounds. All living organisms are composed of organic compounds, as is most of the food you consume, the medicines you take, the fibers in the clothes you wear, and the plastics in the materials you use. Section 2.1 "Chemical Compounds" introduced two organic compounds: methane (CH4) and methanol (CH3OH). These
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and other organic compounds appear frequently in discussions and examples throughout this text. The detection of organic compounds is useful in many fields. In one recently developed application, scientists have devised a new method called “material degradomics” to make it possible to monitor the degradation of old books and historical documents. As paper ages, it produces a familiar “old book smell” from the release of organic compounds in gaseous form. The composition of the gas depends on the original type of paper used, a book’s binding, and the applied media. By analyzing these organic gases and isolating the individual components, preservationists are better able to determine the condition of an object and those books and documents most in need of immediate protection. The simplest class of organic compounds is the hydrocarbons26, which consist entirely of carbon and hydrogen. Petroleum and natural gas are complex, naturally occurring mixtures of many different hydrocarbons that furnish raw materials for the chemical industry. The four major classes of hydrocarbons are the alkanes27, which contain only carbon–hydrogen and carbon–carbon single bonds; the alkenes28, which contain at least one carbon–carbon double bond; the alkynes29, which contain at least one carbon–carbon triple bond; and the aromatic hydrocarbons30, which usually contain rings of six carbon atoms that can be drawn with alternating single and double bonds. Alkanes are also called saturated hydrocarbons, whereas hydrocarbons that contain multiple bonds (alkenes, alkynes, and aromatics) are unsaturated.
Alkanes
26. The simplest class of organic molecules, consisting of only carbon and hydrogen. 27. A saturated hydrocarbon with only carbon–hydrogen and carbon–carbon single bonds. 28. An unsaturated hydrocarbon with at least one carbon–carbon double bond.
The simplest alkane is methane (CH4), a colorless, odorless gas that is the major component of natural gas. In larger alkanes whose carbon atoms are joined in an unbranched chain (straight-chain alkanes), each carbon atom is bonded to at most two other carbon atoms. The structures of two simple alkanes are shown in Figure 2.15 "Straight-Chain Alkanes with Two and Three Carbon Atoms", and the names and condensed structural formulas for the first 10 straight-chain alkanes are in Table 2.7 "The First 10 Straight-Chain Alkanes". The names of all alkanes end in -ane, and their boiling points increase as the number of carbon atoms increases.
29. An unsaturated hydrocarbon with at least one carbon–carbon triple bond. 30. An unsaturated hydrocarbon consisting of a ring of six carbon atoms with alternating single and double bonds.
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Figure 2.15 Straight-Chain Alkanes with Two and Three Carbon Atoms
Table 2.7 The First 10 Straight-Chain Alkanes
Name
2.4 Naming Covalent Compounds
Number of Carbon Atoms
Condensed Structural Formula
Molecular Formula
Boiling Point (°C)
Uses
−162
natural gas constituent
CH3CH3
−89
natural gas constituent
3 C3H 8
CH3CH2CH3
−42 bottled gas
butane
4 C4H10
CH3CH2CH2CH3 or CH3(CH2)2CH3
pentane
5 C5H12
CH3(CH2)3CH3
36
solvent, gasoline
hexane
6 C6H14
CH3(CH2)4CH3
69
solvent, gasoline
heptane
7 C7H16
CH3(CH2)5CH3
98
solvent, gasoline
octane
8 C8H18
CH3(CH2)6CH3
126 gasoline
nonane
9 C9H20
CH3(CH2)7CH3
151 gasoline
decane
10 C10H22
CH3(CH2)8CH3
174 kerosene
methane
1 CH4
CH4
ethane
2 C2H 6
propane
0
lighters, bottled gas
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Alkanes with four or more carbon atoms can have more than one arrangement of atoms. The carbon atoms can form a single unbranched chain, or the primary chain of carbon atoms can have one or more shorter chains that form branches. For example, butane (C4H10) has two possible structures. Normal butane (usually called n-butane) is CH3CH2CH2CH3, in which the carbon atoms form a single unbranched chain. In contrast, the condensed structural formula for isobutane is (CH3)2CHCH3, in which the primary chain of three carbon atoms has a one-carbon chain branching at the central carbon. Three-dimensional representations of both structures are as follows:
The systematic names for branched hydrocarbons use the lowest possible number to indicate the position of the branch along the longest straight carbon chain in the structure. Thus the systematic name for isobutane is 2-methylpropane, which indicates that a methyl group (a branch consisting of –CH 3) is attached to the second carbon of a propane molecule. Similarly, you will learn in Section 2.6 "Industrially Important Chemicals" that one of the major components of gasoline is commonly called isooctane; its structure is as follows:
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As you can see, the compound has a chain of five carbon atoms, so it is a derivative of pentane. There are two methyl group branches at one carbon atom and one methyl group at another. Using the lowest possible numbers for the branches gives 2,2,4-trimethylpentane for the systematic name of this compound.
Alkenes The simplest alkenes are ethylene, C2H4 or CH2=CH2, and propylene, C3H6 or CH3CH=CH2 (part (a) in Figure 2.16 "Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons"). The names of alkenes that have more than three carbon atoms use the same stems as the names of the alkanes (Table 2.7 "The First 10 Straight-Chain Alkanes") but end in -ene instead of -ane. Once again, more than one structure is possible for alkenes with four or more carbon atoms. For example, an alkene with four carbon atoms has three possible structures. One is CH2=CHCH2CH3 (1-butene), which has the double bond between the first and second carbon atoms in the chain. The other two structures have the double bond between the second and third carbon atoms and are forms of CH3CH=CHCH3 (2-butene). All four carbon atoms in 2-butene lie in the same plane, so there are two possible structures (part (a) in Figure 2.16 "Some Simple (a)
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Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons"). If the two methyl groups are on the same side of the double bond, the compound is cis-2-butene (from the Latin cis, meaning “on the same side”). If the two methyl groups are on opposite sides of the double bond, the compound is trans-2-butene (from the Latin trans, meaning “across”). These are distinctly different molecules: cis-2-butene melts at −138.9°C, whereas trans-2-butene melts at −105.5°C. Figure 2.16 Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons
The positions of the carbon atoms in the chain are indicated by C1 or C2.
Just as a number indicates the positions of branches in an alkane, the number in the name of an alkene specifies the position of the first carbon atom of the double bond. The name is based on the lowest possible number starting from either end of the carbon chain, so CH3CH2CH=CH2 is called 1-butene, not 3-butene. Note that CH2=CHCH2CH3 and CH3CH2CH=CH2 are different ways of writing the same molecule (1-butene) in two different orientations.
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The name of a compound does not depend on its orientation. As illustrated for 1-butene, both condensed structural formulas and molecular models show different orientations of the same molecule. Don’t let orientation fool you; you must be able to recognize the same structure no matter what its orientation.
Note the Pattern The positions of groups or multiple bonds are always indicated by the lowest number possible.
Alkynes The simplest alkyne is acetylene, C2H2 or HC≡CH (part (b) in Figure 2.16 "Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons"). Because a mixture of acetylene and oxygen burns with a flame that is hot enough (>3000°C) to cut metals such as hardened steel, acetylene is widely used in cutting and welding torches. The names of other alkynes are similar to those of the corresponding alkanes but end in
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-yne. For example, HC≡CCH3 is propyne, and CH3C≡CCH3 is 2-butyne because the multiple bond begins on the second carbon atom.
Note the Pattern The number of bonds between carbon atoms in a hydrocarbon is indicated in the suffix: • alkane: only carbon–carbon single bonds • alkene: at least one carbon–carbon double bond • alkyne: at least one carbon–carbon triple bond
Cyclic Hydrocarbons In a cyclic hydrocarbon31, the ends of a hydrocarbon chain are connected to form a ring of covalently bonded carbon atoms. Cyclic hydrocarbons are named by attaching the prefix cyclo- to the name of the alkane, the alkene, or the alkyne. The simplest cyclic alkanes are cyclopropane (C3H6) a flammable gas that is also a powerful anesthetic, and cyclobutane (C4H8) (part (c) in Figure 2.16 "Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons"). The most common way to draw the structures of cyclic alkanes is to sketch a polygon with the same number of vertices as there are carbon atoms in the ring; each vertex represents a CH 2 unit. The structures of the cycloalkanes that contain three to six carbon atoms are shown schematically in Figure 2.17 "The Simple Cycloalkanes".
31. A hydrocarbon in which the ends of the carbon chain are connected to form a ring of covalently bonded carbon atoms.
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Figure 2.17 The Simple Cycloalkanes
Aromatic Hydrocarbons Alkanes, alkenes, alkynes, and cyclic hydrocarbons are generally called aliphatic hydrocarbons32. The name comes from the Greek aleiphar, meaning “oil,” because the first examples were extracted from animal fats. In contrast, the first examples of aromatic hydrocarbons, also called arenes, were obtained by the distillation and degradation of highly scented (thus aromatic) resins from tropical trees.
32. Alkanes, alkenes, alkynes, and cyclic hydrocarbons (hydrocarbons that are not aromatic).
The simplest aromatic hydrocarbon is benzene (C6H6), which was first obtained from a coal distillate. The word aromatic now refers to benzene and structurally similar compounds. As shown in part (a) in Figure 2.18 "Two Aromatic Hydrocarbons: (a) Benzene and (b) Toluene", it is possible to draw the structure of benzene in two different but equivalent ways, depending on which carbon atoms are connected by double bonds or single bonds. Toluene is similar to benzene, except that one hydrogen atom is replaced by a –CH3 group; it has the formula C7H8 (part (b) in Figure 2.18 "Two Aromatic Hydrocarbons: (a) Benzene and (b) Toluene"). As you will soon learn, the chemical behavior of aromatic compounds differs from the behavior of aliphatic compounds. Benzene and toluene are found in gasoline, and benzene is the starting material for preparing substances as diverse as aspirin and nylon.
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Figure 2.18 Two Aromatic Hydrocarbons: (a) Benzene and (b) Toluene
Figure 2.19 "Two Hydrocarbons with the Molecular Formula C" illustrates two of the molecular structures possible for hydrocarbons that have six carbon atoms. As you can see, compounds with the same molecular formula can have very different structures. Figure 2.19 Two Hydrocarbons with the Molecular Formula C6H12
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EXAMPLE 10 Write the condensed structural formula for each hydrocarbon. a. b. c. d.
n-heptane 2-pentene 2-butyne cyclooctene
Given: name of hydrocarbon Asked for: condensed structural formula Strategy: A Use the prefix to determine the number of carbon atoms in the molecule and whether it is cyclic. From the suffix, determine whether multiple bonds are present. B Identify the position of any multiple bonds from the number(s) in the name and then write the condensed structural formula. Solution: a. A The prefix hept- tells us that this hydrocarbon has seven carbon atoms, and n- indicates that the carbon atoms form a straight chain. The suffix -ane tells that it is an alkane, with no carbon–carbon double or triple bonds. B The condensed structural formula is CH3CH2CH2CH2CH2CH2CH3, which can also be written as CH3(CH2)5CH3. b. A The prefix pent- tells us that this hydrocarbon has five carbon atoms, and the suffix -ene indicates that it is an alkene, with a carbon–carbon double bond. B The 2- tells us that the double bond begins on the second carbon of the five-carbon atom chain. The condensed structural formula of the compound is therefore CH3CH=CHCH2CH3.
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c. A The prefix but- tells us that the compound has a chain of four carbon atoms, and the suffix -yne indicates that it has a carbon–carbon triple bond. B The 2- tells us that the triple bond begins on the second carbon of the four-carbon atom chain. So the condensed structural formula for the compound is CH3C≡CCH3.
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d. A The prefix cyclo- tells us that this hydrocarbon has a ring structure, and oct- indicates that it contains eight carbon atoms, which we can draw as
The suffix -ene tells us that the compound contains a carbon–carbon double bond, but where in the ring do we place the double bond? B Because all eight carbon atoms are identical, it doesn’t matter. We can draw the structure of cyclooctene as
Exercise Write the condensed structural formula for each hydrocarbon. a. b. c. d.
n-octane 2-hexene 1-heptyne cyclopentane
Answer: a. CH3(CH2)6CH3 b. CH3CH=CHCH2CH2CH3 c. HC≡C(CH2)4CH3 d.
The general name for a group of atoms derived from an alkane is an alkyl group. The name of an alkyl group is derived from the name of the alkane by adding the suffix -yl. Thus the –CH3 fragment is a methyl group, the –CH2CH3 fragment is an ethyl
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group, and so forth, where the dash represents a single bond to some other atom or group. Similarly, groups of atoms derived from aromatic hydrocarbons are aryl groups, which sometimes have unexpected names. For example, the –C 6H5 fragment is derived from benzene, but it is called a phenyl group. In general formulas and structures, alkyl and aryl groups are often abbreviated as R33.
Structures of alkyl and aryl groups. The methyl group is an example of an alkyl group, and the phenyl group is an example of an aryl group.
Alcohols
33. The abbreviation used for alkyl groups and aryl groups in general formulas and structures. 34. A class of organic compounds obtained by replacing one or more of the hydrogen atoms of a hydrocarbon with an −OH group.
Replacing one or more hydrogen atoms of a hydrocarbon with an –OH group gives an alcohol34, represented as ROH. The simplest alcohol (CH3OH) is called either methanol (its systematic name) or methyl alcohol (its common name) (see Figure 2.4 "Different Ways of Representing the Structure of a Molecule"). Methanol is the antifreeze in automobile windshield washer fluids, and it is also used as an efficient fuel for racing cars, most notably in the Indianapolis 500. Ethanol (or ethyl alcohol, CH3CH2OH) is familiar as the alcohol in fermented or distilled beverages, such as beer, wine, and whiskey; it is also used as a gasoline additive (Section 2.6 "Industrially Important Chemicals"). The simplest alcohol derived from an aromatic
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hydrocarbon is C6H5OH, phenol (shortened from phenyl alcohol), a potent disinfectant used in some sore throat medications and mouthwashes. Ethanol, which is easy to obtain from fermentation processes, has successfully been used as an alternative fuel for several decades. Although it is a “green” fuel when derived from plants, it is an imperfect substitute for fossil fuels because it is less efficient than gasoline. Moreover, because ethanol absorbs water from the atmosphere, it can corrode an engine’s seals. Thus other types of processes are being developed that use bacteria to create more complex alcohols, such as octanol, that are more energy efficient and that have a lower tendency to absorb water. As scientists attempt to reduce mankind’s dependence on fossil fuels, the development of these so-called biofuels is a particularly active area of research.
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Summary Covalent inorganic compounds are named by a procedure similar to that used for ionic compounds, using prefixes to indicate the numbers of atoms in the molecular formula. The simplest organic compounds are the hydrocarbons, which contain only carbon and hydrogen. Alkanes contain only carbon–hydrogen and carbon–carbon single bonds, alkenes contain at least one carbon–carbon double bond, and alkynes contain one or more carbon–carbon triple bonds. Hydrocarbons can also be cyclic, with the ends of the chain connected to form a ring. Collectively, alkanes, alkenes, and alkynes are called aliphatic hydrocarbons. Aromatic hydrocarbons, or arenes, are another important class of hydrocarbons that contain rings of carbon atoms related to the structure of benzene (C6H6). A derivative of an alkane or an arene from which one hydrogen atom has been removed is called an alkyl group or an aryl group, respectively. Alcohols are another common class of organic compound, which contain an –OH group covalently bonded to either an alkyl group or an aryl group (often abbreviated R).
KEY TAKEAWAY • Covalent inorganic compounds are named using a procedure similar to that used for ionic compounds, whereas hydrocarbons use a system based on the number of bonds between carbon atoms.
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CONCEPTUAL PROBLEMS 1. Benzene (C6H6) is an organic compound, and KCl is an ionic compound. The sum of the masses of the atoms in each empirical formula is approximately the same. How would you expect the two to compare with regard to each of the following? What species are present in benzene vapor? a. b. c. d.
melting point type of bonding rate of evaporation structure
2. Can an inorganic compound be classified as a hydrocarbon? Why or why not? 3. Is the compound NaHCO3 a hydrocarbon? Why or why not? 4. Name each compound. a. b. c. d. e. f. g.
NiO TiO2 N 2O CS2 SO3 NF3 SF6
5. Name each compound. a. b. c. d. e. f.
HgCl2 IF5 N 2O5 Cl2O HgS PCl5
6. For each structural formula, write the condensed formula and the name of the compound. a.
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b.
c.
d.
e.
7. For each structural formula, write the condensed formula and the name of the compound. a.
b.
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c.
d.
8. Would you expect PCl3 to be an ionic compound or a covalent compound? Explain your reasoning. 9. What distinguishes an aromatic hydrocarbon from an aliphatic hydrocarbon? 10. The following general formulas represent specific classes of hydrocarbons. Refer to Table 2.7 "The First 10 Straight-Chain Alkanes" and Table 2.8 "Some Common Acids That Do Not Contain Oxygen" and Figure 2.16 "Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons" and identify the classes. a. CnH2n + 2 b. CnH2n c. CnH2n − 2 11. Using R to represent an alkyl or aryl group, show the general structure of an a. alcohol. b. phenol.
ANSWER 11. a. ROH (where R is an alkyl group) b. ROH (where R is an aryl group)
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NUMERICAL PROBLEMS 1. Write the formula for each compound. a. b. c. d. e.
dinitrogen monoxide silicon tetrafluoride boron trichloride nitrogen trifluoride phosphorus tribromide
2. Write the formula for each compound. a. b. c. d. e.
dinitrogen trioxide iodine pentafluoride boron tribromide oxygen difluoride arsenic trichloride
3. Write the formula for each compound. a. b. c. d. e.
thallium(I) selenide neptunium(IV) oxide iron(II) sulfide copper(I) cyanide nitrogen trichloride
4. Name each compound. a. b. c. d. e.
RuO4 PbO2 MoF6 Hg2(NO3)2·2H2O WCl4
5. Name each compound. a. b. c. d. e.
NbO2 MoS2 P4S10 Cu2O ReF5
6. Draw the structure of each compound. a. propyne b. ethanol
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c. n-hexane d. cyclopropane e. benzene 7. Draw the structure of each compound. a. b. c. d. e.
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1-butene 2-pentyne cycloheptane toluene phenol
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ANSWERS 1. a. b. c. d. e.
N 2O SiF4 BCl3 NF3 PBr3
3. a. b. c. d. e.
Tl2Se NpO2 FeS CuCN NCl3
5. a. b. c. d. e.
niobium (IV) oxide molybdenum (IV) sulfide tetraphosphorus decasulfide copper(I) oxide rhenium(V) fluoride
7. a. b.
c.
d.
e.
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2.5 Acids and Bases LEARNING OBJECTIVE 1. To identify and name some common acids and bases.
For our purposes at this point in the text, we can define an acid35 as a substance with at least one hydrogen atom that can dissociate to form an anion and an H + ion (a proton) in aqueous solution, thereby forming an acidic solution. We can define bases36 as compounds that produce hydroxide ions (OH−) and a cation when dissolved in water, thus forming a basic solution. Solutions that are neither basic nor acidic are neutral. We will discuss the chemistry of acids and bases in more detail in Chapter 4 "Reactions in Aqueous Solution", Chapter 8 "Ionic versus Covalent Bonding", and Chapter 16 "Aqueous Acid–Base Equilibriums", but in this section we describe the nomenclature of common acids and identify some important bases so that you can recognize them in future discussions. Pure acids and bases and their concentrated aqueous solutions are commonly encountered in the laboratory. They are usually highly corrosive, so they must be handled with care.
Acids
35. A substance with at least one hydrogen atom that can dissociate to form an anion and + an H ion (a proton) in aqueous solution, thereby foming an acidic solution. 36. A substance that produces one or more hydroxide ions (OH− ) and a cation when dissolved in aqueous solution, thereby forming a basic solution.
The names of acids differentiate between (1) acids in which the H + ion is attached to an oxygen atom of a polyatomic anion (these are called oxoacids37, or occasionally oxyacids) and (2) acids in which the H+ ion is attached to some other element. In the latter case, the name of the acid begins with hydro- and ends in -ic, with the root of the name of the other element or ion in between. Recall that the name of the anion derived from this kind of acid always ends in -ide. Thus hydrogen chloride (HCl) gas dissolves in water to form hydrochloric acid (which contains H + and Cl− ions), hydrogen cyanide (HCN) gas forms hydrocyanic acid (which contains H + and CN− ions), and so forth (Table 2.8 "Some Common Acids That Do Not Contain Oxygen"). Examples of this kind of acid are commonly encountered and very important. For instance, your stomach contains a dilute solution of hydrochloric acid to help digest food. When the mechanisms that prevent the stomach from digesting itself malfunction, the acid destroys the lining of the stomach and an ulcer forms.
37. An acid in which the + dissociable H ion is attached to an oxygen atom of a polyatomic anion.
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Note the Pattern Acids are distinguished by whether the H+ ion is attached to an oxygen atom of a polyatomic anion or some other element.
Table 2.8 Some Common Acids That Do Not Contain Oxygen Formula Name in Aqueous Solution Name of Gaseous Species HF
hydrofluoric acid
hydrogen fluoride
HCl
hydrochloric acid
hydrogen chloride
HBr
hydrobromic acid
hydrogen bromide
HI
hydroiodic acid
hydrogen iodide
HCN
hydrocyanic acid
hydrogen cyanide
H 2S
hydrosulfuric acid
hydrogen sulfide
If an acid contains one or more H+ ions attached to oxygen, it is a derivative of one of the common oxoanions, such as sulfate (SO42−) or nitrate (NO3−). These acids contain as many H+ ions as are necessary to balance the negative charge on the anion, resulting in a neutral species such as H2SO4 and HNO3. The names of acids are derived from the names of anions according to the following rules: 1. If the name of the anion ends in -ate, then the name of the acid ends in -ic. For example, because NO3− is the nitrate ion, HNO3 is nitric acid. Similarly, ClO4− is the perchlorate ion, so HClO4 is perchloric acid. Two important acids are sulfuric acid (H2SO4) from the sulfate ion (SO42−) and phosphoric acid (H3PO4) from the phosphate ion (PO43−). These two names use a slight variant of the root of the anion name: sulfate becomes sulfuric and phosphate becomes phosphoric. 2. If the name of the anion ends in -ite, then the name of the acid ends in -ous. For example, OCl− is the hypochlorite ion, and HOCl is hypochlorous acid; NO2− is the nitrite ion, and HNO2 is nitrous acid; and SO32− is the sulfite ion, and H2SO3 is sulfurous acid. The same roots
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are used whether the acid name ends in -ic or -ous; thus, sulfite becomes sulfurous. The relationship between the names of the oxoacids and the parent oxoanions is illustrated in Figure 2.20 "The Relationship between the Names of the Oxoacids and the Names of the Parent Oxoanions", and some common oxoacids are in Table 2.9 "Some Common Oxoacids". Figure 2.20 The Relationship between the Names of the Oxoacids and the Names of the Parent Oxoanions
Table 2.9 Some Common Oxoacids Formula
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Name
HNO2
nitrous acid
HNO3
nitric acid
H2SO3
sulfurous acid
H2SO4
sulfuric acid
H3PO4
phosphoric acid
H2CO3
carbonic acid
HClO
hypochlorous acid
HClO2
chlorous acid
HClO3
chloric acid
HClO4
perchloric acid
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EXAMPLE 11 Name and give the formula for each acid. a. the acid formed by adding a proton to the hypobromite ion (OBr −) b. the acid formed by adding two protons to the selenate ion (SeO 42−) Given: anion Asked for: parent acid Strategy: Refer to Table 2.8 "Some Common Acids That Do Not Contain Oxygen" and Table 2.9 "Some Common Oxoacids" to find the name of the acid. If the acid is not listed, use the guidelines given previously. Solution: Neither species is listed in Table 2.8 "Some Common Acids That Do Not Contain Oxygen" or Table 2.9 "Some Common Oxoacids", so we must use the information given previously to derive the name of the acid from the name of the polyatomic anion. a. The anion name, hypobromite, ends in -ite, so the name of the parent acid ends in -ous. The acid is therefore hypobromous acid (HOBr). b. Selenate ends in -ate, so the name of the parent acid ends in -ic. The acid is therefore selenic acid (H2SeO4). Exercise Name and give the formula for each acid. a. the acid formed by adding a proton to the perbromate ion (BrO 4−) b. the acid formed by adding three protons to the arsenite ion (AsO 33−) Answer: a. perbromic acid; HBrO4 b. arsenous acid; H3AsO3
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Many organic compounds contain the carbonyl group38, in which there is a carbon–oxygen double bond. In carboxylic acids39, an –OH group is covalently bonded to the carbon atom of the carbonyl group. Their general formula is RCO 2H, sometimes written as RCOOH:
where R can be an alkyl group, an aryl group, or a hydrogen atom. The simplest example, HCO2H, is formic acid, so called because it is found in the secretions of stinging ants (from the Latin formica, meaning “ant”). Another example is acetic acid (CH3CO2H), which is found in vinegar. Like many acids, carboxylic acids tend to have sharp odors. For example, butyric acid (CH3CH2CH2CO2H), is responsible for the smell of rancid butter, and the characteristic odor of sour milk and vomit is due to lactic acid [CH3CH(OH)CO2H]. Some common carboxylic acids are shown in Figure 2.21 "Some Common Carboxylic Acids".
38. A carbon atom double-bonded to an oxygen atom. It is a characteristic feature of many organic compounds, including carboxylic acids. 39. An organic compound that contains an −OH group covalently bonded to the carbon atom of a carbonyl group. The general formula of a carboxylic acid is RCO 2 H. In water a carboxylic acid dissociates to produce an acidic solution.
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Figure 2.21 Some Common Carboxylic Acids
Although carboxylic acids are covalent compounds, when they dissolve in water, they dissociate to produce H+ ions (just like any other acid) and RCO2− ions. Note that only the hydrogen attached to the oxygen atom of the CO2 group dissociates to form an H+ ion. In contrast, the hydrogen atom attached to the oxygen atom of an alcohol does not dissociate to form an H+ ion when an alcohol is dissolved in water. The reasons for the difference in behavior between carboxylic acids and alcohols will be discussed in Chapter 8 "Ionic versus Covalent Bonding".
Note the Pattern Only the hydrogen attached to the oxygen atom of the CO2 group dissociates to form an H+ ion.
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Bases We will present more comprehensive definitions of bases in later chapters, but virtually every base you encounter in the meantime will be an ionic compound, such as sodium hydroxide (NaOH) and barium hydroxide [Ba(OH) 2], that contain the hydroxide ion and a metal cation. These have the general formula M(OH) n. It is important to recognize that alcohols, with the general formula ROH, are covalent compounds, not ionic compounds; consequently, they do not dissociate in water to form a basic solution (containing OH− ions). When a base reacts with any of the acids we have discussed, it accepts a proton (H+). For example, the hydroxide ion (OH−) accepts a proton to form H2O. Thus bases are also referred to as proton acceptors. Concentrated aqueous solutions of ammonia (NH3) contain significant amounts of the hydroxide ion, even though the dissolved substance is not primarily ammonium hydroxide (NH4OH) as is often stated on the label. Thus aqueous ammonia solution is also a common base. Replacing a hydrogen atom of NH3 with an alkyl group results in an amine40 (RNH2), which is also a base. Amines have pungent odors—for example, methylamine (CH3NH2) is one of the compounds responsible for the foul odor associated with spoiled fish. The physiological importance of amines is suggested in the word vitamin, which is derived from the phrase vital amines. The word was coined to describe dietary substances that were effective at preventing scurvy, rickets, and other diseases because these substances were assumed to be amines. Subsequently, some vitamins have indeed been confirmed to be amines.
Note the Pattern Metal hydroxides (MOH) yield OH− ions and are bases, alcohols (ROH) do not yield OH− or H+ ions and are neutral, and carboxylic acids (RCO2H) yield H+ ions and are acids.
40. An organic compound that has the general formula RNH2 , where R is an alkyl group. Amines, like ammonia, are bases.
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Summary Common acids and the polyatomic anions derived from them have their own names and rules for nomenclature. The nomenclature of acids differentiates between oxoacids, in which the H+ ion is attached to an oxygen atom of a polyatomic ion, and acids in which the H+ ion is attached to another element. Carboxylic acids are an important class of organic acids. Ammonia is an important base, as are its organic derivatives, the amines.
KEY TAKEAWAY • Common acids and polyatomic anions derived from them have their own names and rules for nomenclature.
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CONCEPTUAL PROBLEMS 1. Name each acid. a. b. c. d. e.
HCl HBrO3 HNO3 H2SO4 HIO3
2. Name each acid. a. b. c. d. e.
HBr H2SO3 HClO3 HCN H3PO4
3. Name the aqueous acid that corresponds to each gaseous species. a. hydrogen bromide b. hydrogen cyanide c. hydrogen iodide 4. For each structural formula, write the condensed formula and the name of the compound. a.
b.
5. For each structural formula, write the condensed formula and the name of the compound. a.
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b.
6. When each compound is added to water, is the resulting solution acidic, neutral, or basic? a. b. c. d. e. f.
CH3CH2OH Mg(OH)2 C6H5CO2H LiOH C3H7CO2H H2SO4
7. Draw the structure of the simplest example of each type of compound. a. b. c. d. e. f. g. h.
alkane alkene alkyne aromatic hydrocarbon alcohol carboxylic acid amine cycloalkane
8. Identify the class of organic compound represented by each compound. a.
b. CH3CH2OH c. HC≡CH d.
e. C3H7NH2
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f. CH3CH=CHCH2CH3 g.
h.
9. Identify the class of organic compound represented by each compound. a.
b. c.
d.
e.
f. CH3C≡CH g.
h.
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NUMERICAL PROBLEMS 1. Write the formula for each compound. a. b. c. d. e.
hypochlorous acid perbromic acid hydrobromic acid sulfurous acid sodium perbromate
2. Write the formula for each compound. a. b. c. d. e.
hydroiodic acid hydrogen sulfide phosphorous acid perchloric acid calcium hypobromite
3. Name each compound. a. b. c. d. e.
HBr H2SO3 HCN HClO4 NaHSO4
4. Name each compound. a. b. c. d. e.
2.5 Acids and Bases
H2SO4 HNO2 K2HPO4 H3PO3 Ca(H2PO4)2·H2O
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2.6 Industrially Important Chemicals LEARNING OBJECTIVE 1. To appreciate the scope of the chemical industry and its contributions to modern society.
It isn’t easy to comprehend the scale on which the chemical industry must operate to supply the huge amounts of chemicals required in modern industrial societies. Figure 2.22 "Top 25 Chemicals Produced in the United States in 2002*" lists the names and formulas of the chemical industry’s “top 25” for 2002—the 25 chemicals produced in the largest quantity in the United States that year—along with the amounts produced, in billions of pounds. To put these numbers in perspective, consider that the 88.80 billion pounds of sulfuric acid produced in the United States in 2002 has a volume of 21.90 million cubic meters (2.19 × 107 m3), enough to fill the Pentagon, probably the largest office building in the world, about 22 times. Figure 2.22 Top 25 Chemicals Produced in the United States in 2002*
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According to Figure 2.22 "Top 25 Chemicals Produced in the United States in 2002*", 11 of the top 15 compounds produced in the United States are inorganic, and the total mass of inorganic chemicals produced is almost twice the mass of organic chemicals. Yet the diversity of organic compounds used in industry is such that over half of the top 25 compounds (13 out of 25) are organic. Why are such huge quantities of chemical compounds produced annually? They are used both directly as components of compounds and materials that we encounter on an almost daily basis and indirectly in the production of those compounds and materials. The single largest use of industrial chemicals is in the production of foods: 7 of the top 15 chemicals are either fertilizers (ammonia, urea, and ammonium nitrate) or used primarily in the production of fertilizers (sulfuric acid, nitric acid, nitrogen, and phosphoric acid). Many of the organic chemicals on the list are used primarily as ingredients in the plastics and related materials that are so prevalent in contemporary society. Ethylene and propylene, for example, are used to produce polyethylene and polypropylene, which are made into plastic milk bottles, sandwich bags, indoor-outdoor carpets, and other common items. Vinyl chloride, in the form of polyvinylchloride, is used in everything from pipes to floor tiles to trash bags. Though not listed in Figure 2.22 "Top 25 Chemicals Produced in the United States in 2002*", butadiene and carbon black are used in the manufacture of synthetic rubber for tires, and phenol and formaldehyde are ingredients in plywood, fiberglass, and many hard plastic items. We do not have the space in this text to consider the applications of all these compounds in any detail, but we will return to many of them after we have developed the concepts necessary to understand their underlying chemistry. Instead, we conclude this chapter with a brief discussion of petroleum refining as it relates to gasoline and octane ratings and a look at the production and use of the topmost industrial chemical, sulfuric acid.
Petroleum The petroleum that is pumped out of the ground at locations around the world is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundances of the components vary depending on the source. So Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules.
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Gasoline Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall from Chapter 1 "Introduction to Chemistry" that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in Figure 2.23 "The Distillation of Petroleum" shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F), at which temperature it has become a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in Figure 2.23 "The Distillation of Petroleum" shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs. Figure 2.23 The Distillation of Petroleum
(a) This is a diagram of a distillation column used for separating petroleum fractions. (b) Petroleum fractions condense at different temperatures, depending on the number of carbon atoms in the molecules, and are drawn off from the column. The most volatile components (those with the lowest boiling points) condense at the top of the column, and the least volatile (those with the highest boiling points) condense at the bottom.
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The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions must be converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking41, in which the larger and heavier hydrocarbons in the kerosene and higher-boilingpoint fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming42; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Using metals such as platinum brings about the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation.
Octane Ratings
41. A process in petroleum refining in which the larger and heavier hydrocarbons in kerosene and higher-boilingpoint fractions are heated to high temperatures, causing the carbon–carbon bonds to break (“crack”), thus producing a more volatile mixture. 42. The second process used in petroleum refining, which is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons.
The quality of a fuel is indicated by its octane rating43, which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion (Figure 2.24 "The Burning of Gasoline in an Internal Combustion Engine"), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level.
43. A measure of a fuel’s ability to burn in a combustion engine without knocking or pinging (indications of premature combustion). The higher the octane rating, the higher quality the fuel.
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Figure 2.24 The Burning of Gasoline in an Internal Combustion Engine
(a) Normally, fuel is ignited by the spark plug, and combustion spreads uniformly outward. (b) Gasoline with an octane rating that is too low for the engine can ignite prematurely, resulting in uneven burning that causes knocking and pinging.
The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture: 0.89(100) + 0.11(0) = 89 A gasoline that performs at the same level as a blend of 89% isooctane and 11% nheptane is assigned an octane rating of 89; this represents an intermediate grade of gasoline. Regular gasoline typically has an octane rating of 87; premium has a rating of 93 or higher. As shown in Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives", many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition,
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antiknock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyllead [(C 2H5)4Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place. They combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing. Figure 2.25 The Octane Ratings of Some Hydrocarbons and Common Additives
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EXAMPLE 12 You have a crude (i.e., unprocessed or straight-run) petroleum distillate consisting of 10% n-heptane, 10% n-hexane, and 80% n-pentane by mass, with an octane rating of 52. What percentage of MTBE by mass would you need to increase the octane rating of the distillate to that of regular-grade gasoline (a rating of 87), assuming that the octane rating is directly proportional to the amounts of the compounds present? Use the information presented in Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives". Given: composition of petroleum distillate, initial octane rating, and final octane rating Asked for: percentage of MTBE by mass in final mixture Strategy: A Define the unknown as the percentage of MTBE in the final mixture. Then subtract this unknown from 100% to obtain the percentage of petroleum distillate. B Multiply the percentage of MTBE and the percentage of petroleum distillate by their respective octane ratings; add these values to obtain the overall octane rating of the new mixture. C Solve for the unknown to obtain the percentage of MTBE needed. Solution: A The question asks what percentage of MTBE will give an overall octane rating of 87 when mixed with the straight-run fraction. From Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives", the octane rating of MTBE is 116. Let x be the percentage of MTBE, and let 100 − x be the percentage of petroleum distillate. B Multiplying the percentage of each component by its respective octane rating and setting the sum equal to the desired octane rating of the mixture (87) times 100 gives
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final octane rating of mixture
= 87(100)
= 52(100 − x) + 116x = 5200 − 52x + 116x = 5200 + 64x
C Solving the equation gives x = 55%. Thus the final mixture must contain 55% MTBE by mass. To obtain a composition of 55% MTBE by mass, you would have to add more than an equal mass of MTBE (actually 0.55/0.45, or 1.2 times) to the straightrun fraction. This is 1.2 tons of MTBE per ton of straight-run gasoline, which would be prohibitively expensive. Thus there are sound economic reasons for reforming the kerosene fractions to produce toluene and other aromatic compounds, which have high octane ratings and are much cheaper than MTBE. Exercise As shown in Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives", toluene is one of the fuels suitable for use in automobile engines. How much toluene would have to be added to a blend of the petroleum fraction in this example containing 15% MTBE by mass to increase the octane rating to that of premium gasoline (93)? Answer: The final blend is 56% toluene by mass, which requires a ratio of 56/44, or 1.3 tons of toluene per ton of blend.
Sulfuric Acid Sulfuric acid is one of the oldest chemical compounds known. It was probably first prepared by alchemists who burned sulfate salts such as FeSO 4·7H2O, called green vitriol from its color and glassy appearance (from the Latin vitrum, meaning “glass”). Because pure sulfuric acid was found to be useful for dyeing textiles, enterprising individuals looked for ways to improve its production. By the mid-18th century, sulfuric acid was being produced in multiton quantities by the lead-chamber process, which was invented by John Roebuck in 1746. In this process, sulfur was burned in a large room lined with lead, and the resulting fumes were absorbed in water.
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Production The production of sulfuric acid today is likely to start with elemental sulfur obtained through an ingenious technique called the Frasch process, which takes advantage of the low melting point of elemental sulfur (115.2°C). Large deposits of elemental sulfur are found in porous limestone rocks in the same geological formations that often contain petroleum. In the Frasch process, water at high temperature (160°C) and high pressure is pumped underground to melt the sulfur, and compressed air is used to force the liquid sulfur-water mixture to the surface (Figure 2.26 "Extraction of Elemental Sulfur from Underground Deposits"). The material that emerges from the ground is more than 99% pure sulfur. After it solidifies, it is pulverized and shipped in railroad cars to the plants that produce sulfuric acid, as shown here. Figure 2.26 Extraction of Elemental Sulfur from Underground Deposits
Transporting sulfur. A train carries elemental sulfur through the White Canyon of the Thompson River in British Columbia, Canada.
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In the Frasch process for extracting sulfur, very hot water at high pressure is injected into the sulfur-containing rock layer to melt the sulfur. The resulting mixture of liquid sulfur and hot water is forced up to the surface by compressed air.
An increasing number of sulfuric acid manufacturers have begun to use sulfur dioxide (SO2) as a starting material instead of elemental sulfur. Sulfur dioxide is recovered from the burning of oil and gas, which contain small amounts of sulfur compounds. When not recovered, SO2 is released into the atmosphere, where it is converted to an environmentally hazardous form that leads to acid rain (Chapter 4 "Reactions in Aqueous Solution"). If sulfur is the starting material, the first step in the production of sulfuric acid is the combustion of sulfur with oxygen to produce SO2. Next, SO2 is converted to SO3 by the contact process, in which SO2 and O2 react in the presence of V2O5 to achieve about 97% conversion to SO3. The SO3 can then be treated with a small amount of water to produce sulfuric acid. Usually, however, the SO3 is absorbed in concentrated sulfuric acid to produce oleum, a more potent form called fuming sulfuric acid. Because of its high SO3 content (approximately 99% by mass), oleum is cheaper to ship than concentrated sulfuric acid. At the point of use, the oleum is diluted with water to give concentrated sulfuric acid (very carefully because dilution generates enormous amounts of heat). Because SO2 is a pollutant, the small amounts of unconverted SO2 are recovered and recycled to minimize the amount released into the air.
Uses Two-thirds of the sulfuric acid produced in the United States is used to make fertilizers, most of which contain nitrogen, phosphorus, and potassium (in a form called potash). In earlier days, phosphate-containing rocks were simply ground up and spread on fields as fertilizer, but the extreme insolubility of many salts that contain the phosphate ion (PO43−) limits the availability of phosphorus from these sources. Sulfuric acid serves as a source of protons (H+ ions) that react with phosphate minerals to produce more soluble salts containing HPO 42− or H2PO4− as the anion, which are much more readily taken up by plants. In this context, sulfuric acid is used in two principal ways: (1) the phosphate rocks are treated with concentrated sulfuric acid to produce “superphosphate,” a mixture of 32% CaHPO 4 and Ca(H2PO4)2·H2O, 50% CaSO4·2H2O, approximately 3% absorbed phosphoric acid, and other nutrients; and (2) sulfuric acid is used to produce phosphoric acid (H3PO4), which can then be used to convert phosphate rocks to “triple superphosphate,” which is largely Ca(H2PO4)2·H2O.
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Sulfuric acid is also used to produce potash, one of the other major ingredients in fertilizers. The name potash originally referred to potassium carbonate (obtained by boiling wood ashes with water in iron pots), but today it also refers to compounds such as potassium hydroxide (KOH) and potassium oxide (K 2O). The usual source of potassium in fertilizers is actually potassium sulfate (K 2SO4), which is produced by several routes, including the reaction of concentrated sulfuric acid with solid potassium chloride (KCl), which is obtained as the pure salt from mineral deposits.
Summary Many chemical compounds are prepared industrially in huge quantities and used to produce foods, fuels, plastics, and other such materials. Petroleum refining takes a complex mixture of naturally occurring hydrocarbons as a feedstock and, through a series of steps involving distillation, cracking, and reforming, converts them to mixtures of simpler organic compounds with desirable properties. A major use of petroleum is in the production of motor fuels such as gasoline. The performance of such fuels in engines is described by their octane rating, which depends on the identity of the compounds present and their relative abundance in the blend. Sulfuric acid is the compound produced in the largest quantity in the industrial world. Much of the sulfur used in the production of sulfuric acid is obtained via the Frasch process, in which very hot water forces liquid sulfur out of the ground in nearly pure form. Sulfuric acid is produced by the reaction of sulfur dioxide with oxygen in the presence of vanadium(V) oxide (the contact process), followed by absorption of the sulfur trioxide in concentrated sulfuric acid to produce oleum. Most sulfuric acid is used to prepare fertilizers.
KEY TAKEAWAY • Many chemical compounds are prepared industrially in huge quantities to prepare the materials we need and use in our daily lives.
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CONCEPTUAL PROBLEMS 1. Describe the processes used for converting crude oil to transportation fuels. 2. If your automobile engine is knocking, is the octane rating of your gasoline too low or too high? Explain your answer. 3. Tetraethyllead is no longer used as a fuel additive to prevent knocking. Instead, fuel is now marketed as “unleaded.” Why is tetraethyllead no longer used? 4. If you were to try to extract sulfur from an underground source, what process would you use? Describe briefly the essential features of this process. 5. Why are phosphate-containing minerals used in fertilizers treated with sulfuric acid?
ANSWER 5. Phosphate salts contain the highly-charged PO43− ion, salts of which are often insoluble. Protonation of the PO43− ion by strong acids such as H2SO4 leads to the formation of the HPO42− and H2PO4− ions. Because of their decreased negative charge, salts containing these anions are usually much more soluble, allowing the anions to be readily taken up by plants when they are applied as fertilizer.
NUMERICAL PROBLEM 1. In Example 12, the crude petroleum had an overall octane rating of 52. What is the composition of a solution of MTBE and n-heptane that has this octane rating?
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2.7 End-of-Chapter Material
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APPLICATION PROBLEMS Problems marked with a ♦ involve multiple concepts. 1. Carbon tetrachloride (CCl4) was used as a dry cleaning solvent until it was found to cause liver cancer. Based on the structure of chloroform given in Section 2.1 "Chemical Compounds", draw the structure of carbon tetrachloride. 2. Ammonium nitrate and ammonium sulfate are used in fertilizers as a source of nitrogen. The ammonium cation is tetrahedral. Refer to Section 2.1 "Chemical Compounds" to draw the structure of the ammonium ion. 3. The white light in fireworks displays is produced by burning magnesium in air, which contains oxygen. What compound is formed? 4. Sodium hydrogen sulfite, which is used for bleaching and swelling leather and to preserve flavor in almost all commercial wines, is made from sulfur dioxide. What are the formulas for these two sulfur-containing compounds? 5. Carbonic acid is used in carbonated drinks. When combined with lithium hydroxide, it produces lithium carbonate, a compound used to increase the brightness of pottery glazes and as a primary treatment for depression and bipolar disorder. Write the formula for both of these carbon-containing compounds. 6. Vinegar is a dilute solution of acetic acid, an organic acid, in water. What grouping of atoms would you expect to find in the structural formula for acetic acid? 7. ♦ Sodamide, or sodium amide, is prepared from sodium metal and gaseous ammonia. Sodamide contains the amide ion (NH2−), which reacts with water to form the hydroxide anion by removing an H+ ion from water. Sodium amide is also used to prepare sodium cyanide. a. Write the formula for each of these sodium-containing compounds. b. What are the products of the reaction of sodamide with water? 8. A mixture of isooctane, n-pentane, and n-heptane is known to have an octane rating of 87. Use the data in Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives" to calculate how much isooctane and nheptane are present if the mixture is known to contain 30% n-pentane. 9. A crude petroleum distillate consists of 60% n-pentane, 25% methanol, and the remainder n-hexane by mass (Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives").
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a. What is the octane rating? b. How much MTBE would have to be added to increase the octane rating to 93? 10. Premium gasoline sold in much of the central United States has an octane rating of 93 and contains 10% ethanol. What is the octane rating of the gasoline fraction before ethanol is added? (See Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives".)
ANSWERS
1. 3. MgO, magnesium oxide 5. Carbonic acid is H2CO3; lithium carbonate is Li2CO3. 7. a. Sodamide is NaNH2, and sodium cyanide is NaCN. b. Sodium hydroxide (NaOH) and ammonia (NH3). 9. a. 68 b. 52 g of MTBE must be added to 48 g of the crude distillate.
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Chapter 3 Chemical Reactions Chapter 2 "Molecules, Ions, and Chemical Formulas" introduced you to a wide variety of chemical compounds, many of which have interesting applications. For example, nitrous oxide, a mild anesthetic, is also used as the propellant in cans of whipped cream, while copper(I) oxide is used as both a red glaze for ceramics and in antifouling bottom paints for boats. In addition to the physical properties of substances, chemists are also interested in their chemical reactions1, processes in which a substance is converted to one or more other substances with different compositions and properties. Our very existence depends on chemical reactions, such as those between oxygen in the air we breathe and nutrient molecules in the foods we eat. Other reactions cook those foods, heat our homes, and provide the energy to run our cars. Many of the materials and pharmaceuticals that we take for granted today, such as silicon nitride for the sharp edge of cutting tools and antibiotics such as amoxicillin, were unknown only a few years ago. Their development required that chemists understand how substances combine in certain ratios and under specific conditions to produce a new substance with particular properties.
1. A process in which a substance is converted to one or more other substances with different compositions and properties.
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Sodium. The fourth most abundant alkali metal on Earth, sodium is a highly reactive element that is never found free in nature. When heated to 250°C, it bursts into flames if exposed to air.
We begin this chapter by describing the relationship between the mass of a sample of a substance and its composition. We then develop methods for determining the quantities of compounds produced or consumed in chemical reactions, and we describe some fundamental types of chemical reactions. By applying the concepts and skills introduced in this chapter, you will be able to explain what happens to the sugar in a candy bar you eat, what reaction occurs in a battery when you start your car, what may be causing the “ozone hole” over Antarctica, and how we might prevent the hole’s growth.
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3.1 The Mole and Molar Masses LEARNING OBJECTIVE 1. To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound and to calculate the number of atoms, molecules, or formula units in a sample of a substance.
As you learned in Chapter 1 "Introduction to Chemistry", the mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu). Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, we can calculate the average atomic mass of any molecule or polyatomic ion from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations.
Molecular and Formula Masses The molecular mass2 of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example 1.
2. The sum of the average masses of the atoms in one molecule of a substance, each multiplied by its subscript.
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EXAMPLE 1 Calculate the molecular mass of ethanol, whose condensed structural formula is CH3CH2OH. Among its many uses, ethanol is a fuel for internal combustion engines. Given: molecule Asked for: molecular mass Strategy: A Determine the number of atoms of each element in the molecule. B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element. C Add together the masses to give the molecular mass. Solution: A The molecular formula of ethanol may be written in three different ways: CH3CH2OH (which illustrates the presence of an ethyl group, CH 3CH2−, and an −OH group), C2H5OH, and C2H6O; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom. B Taking the atomic masses from the periodic table, we obtain
2 × atomic mass of carbon = 2 atoms
(
6 × atomic mass of hydrogen = 6 atoms 1 × atomic mass of oxygen = 1 atom
(
12.011 amu atom (
)
1.0079 amu atom
15.9994 amu atom
= 24.022 amu )
)
= 6.0474 am
= 15.9994 am
C Adding together the masses gives the molecular mass:
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24.022 amu + 6.0474 amu + 15.9994 amu = 46.069 amu Alternatively, we could have used unit conversions to reach the result in one step, as described in Essential Skills 2 (Section 3.7 "Essential Skills 2"):
[
2 atoms C
( 1 atom C )] 12.011 amu
+
[
6 atoms H
( 1 atom H )] 1.0079 amu
The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
2C 6H +1O C2 H6 O
(2 atoms)(12.011 amu/atom) (6 atoms)(1.0079 amu/atom) (1 atom)(15.9994 amu/atom) molecular mass of ethanol
= = = =
24.022 amu 6.0474 amu 15.9994 amu 46.069 amu
Exercise Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is CCl3F. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:
Answer: 137.368 amu
3. Another name for formula mass. 4. The sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript.
3.1 The Mole and Molar Masses
Unlike molecules, which have covalent bonds, ionic compounds do not have a readily identifiable molecular unit. So for ionic compounds we use the formula mass (also called the empirical formula mass3) of the compound rather than the molecular mass. The formula mass4 is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or
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+
[
1
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implied). It is directly analogous to the molecular mass of a covalent compound. Once again, the units are atomic mass units.
Note the Pattern Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
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EXAMPLE 2 Calculate the formula mass of Ca3(PO4)2, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk. Given: ionic compound Asked for: formula mass Strategy: A Determine the number of atoms of each element in the empirical formula. B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element. C Add together the masses to give the formula mass. Solution: A The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43− ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms. B Taking atomic masses from the periodic table, we obtain
3 × atomic mass of calcium = 3 atoms
(
2 × atomic mass of phosphorus = 2 atoms 8 × atomic mass of oxygen = 8 atoms
3.1 The Mole and Molar Masses
(
40.078 amu atom (
)
= 120.234 am
30.973761 amu atom
15.9994 amu atom
)
)
= 61
= 127.9952
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C Adding together the masses gives the formula mass of Ca 3(PO4)2: 120.234 amu + 61.947522 amu + 127.9952 amu = 310.177 amu We could also find the formula mass of Ca3(PO4)2 in one step by using unit conversions or a tabular format:
[
3 atoms Ca
3Ca 2P +8O Ca3 P2 O8
( 1 atom Ca )] 40.078 amu
+
[
2 atoms P
(3 atoms)(40.078 amu/atom) (2 atoms)(30.973761 amu/atom) (8 atoms)(15.9994 amu/atom) formula mass of Ca 3 (PO4 )2
= = = =
(
30.973761 amu 1 atom P 120.234 amu 61.947522 amu 127.9952 amu 310.177 amu
Exercise Calculate the formula mass of Si3N4, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys. Answer: 140.29 amu
The Mole
5. A process in which a substance is converted to one or more other substances with different compositions and properties.
3.1 The Mole and Molar Masses
In Chapter 1 "Introduction to Chemistry", we described Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. We also described the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10 −23 g/ atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction5, it is therefore absolutely essential for chemists to know how many atoms or molecules are contained in a
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)]
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measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol)6, from the Latin moles, meaning “pile” or “heap” (not from the small subterranean animal!). Many familiar items are sold in numerical quantities that have unusual names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extraordinarily large numerical unit is needed to count them. The mole is used for this purpose. A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 1023 atoms, but for most purposes 6.022 × 1023 provides an adequate number of significant figures. Just as 1 mol of atoms contains 6.022 × 1023 atoms, 1 mol of eggs contains 6.022 × 1023 eggs. The number in a mole is called Avogadro’s number7, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain. It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mol of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 1023.
Note the Pattern One mole always has the same number of objects: 6.022 × 1023. 6. The quantity of a substance that contains the same number of units (e.g., atoms or molecules) as the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. 7. The number of units (e.g., atoms, molecules, or formula units) in 1 mol:
6.022 × 10 23 .
3.1 The Mole and Molar Masses
To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 1017 mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would get more than one trillion dollars. Clearly, the mole is so large that it is useful only for measuring very small objects, such as atoms.
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The concept of the mole allows us to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, we would weigh out 12 g of isotopically pure carbon-12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 1023). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way in which the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, we can now restate Dalton’s theory: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H2O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.
Molar Mass The molar mass8 of a substance is defined as the mass in grams of 1 mol of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.
Note the Pattern The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
8. The mass in grams of 1 mol of a substance.
3.1 The Mole and Molar Masses
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:
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Substance (formula) carbon (C)
Atomic, Molecular, or Formula Mass (amu)
Molar Mass (g/mol)
12.011 (atomic mass)
12.011
ethanol (C2H5OH)
46.069 (molecular mass)
46.069
calcium phosphate [Ca3(PO4)2]
310.177 (formula mass)
310.177
The molar mass of naturally occurring carbon is different from that of carbon-12 and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 10 12) are carbon-14. (For more information, see Section 1.6 "Isotopes and Atomic Masses".) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When we deal with elements such as iodine and sulfur, which occur as a diatomic molecule (I2) and a polyatomic molecule (S8), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I2 and S8). The molar mass of ethanol is the mass of ethanol (C 2H5OH) that contains 6.022 × 1023 ethanol molecules. As you calculated in Example 1, the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca3(PO4)2] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 1023 formula units. Figure 3.1 "Samples of 1 Mol of Some Common Substances" shows samples that contain precisely one molar mass of several common substances. The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, we use the relationship
3.1 The Mole and Molar Masses
Figure 3.1 Samples of 1 Mol of Some Common Substances
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Equation 3.1 (moles)(molar mass) → mass or, more specifically,
moles
( mole ) grams
= grams
Conversely, to convert the mass of a substance to moles, we use Equation 3.2
mass → moles ( molar mass ) grams mole = grams = moles ( grams/mole ) ( grams ) Be sure to pay attention to the units when converting between mass and moles. Figure 3.2 "A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units" is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example 3 and Example 4. Figure 3.2 A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units
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EXAMPLE 3 For 35.00 g of ethylene glycol (HOCH2CH2OH), which is used in inks for ballpoint pens, calculate the number of a. moles. b. molecules. Given: mass and molecular formula Asked for: number of moles and number of molecules Strategy: A Use the molecular formula of the compound to calculate its molecular mass in grams per mole. B Convert from mass to moles by dividing the mass given by the compound’s molar mass. C Convert from moles to molecules by multiplying the number of moles by Avogadro’s number. Solution:
a. A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example 1:
2C 6H +2O C2 H6 O2
(2 atoms)(12.011 amu/atom) (6 atoms)(1.0079 amu/atom) (2 atoms)(15.9994 amu/atom) molecular mass of ethylene glycol
= = = =
24.022 amu 6.0474 amu 31.9988 amu 62.068 amu
The molar mass of ethylene glycol is 62.068 g/mol. B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):
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mass of ethylene glycol (g) = moles ethylene glycol (mol) molar mass (g/mol) So
1 mol ethylene glycol 35.00 g ethylene glycol 62.068 g ethylene glycol
= 0.5639 m
It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations. b. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:
molecules of ethylene glycol
= 0.5639 mol
(
6.022 × 10 23 mo 1 mol
= 3.396 × 10 23 molecules Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 1023 molecules, which is indeed the case. Exercise For 75.0 g of CCl3F (Freon-11), calculate the number of a. moles. b. molecules. Answer: a. 0.546 mol b. 3.29 × 1023 molecules
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EXAMPLE 4 Calculate the mass of 1.75 mol of each compound. a. S2Cl2 (common name: sulfur monochloride; systematic name: disulfur dichloride) b. Ca(ClO)2 (calcium hypochlorite) Given: number of moles and molecular or empirical formula Asked for: mass Strategy: A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic). B Convert from moles to mass by multiplying the moles of the compound given by its molar mass. Solution: We begin by calculating the molecular mass of S2Cl2 and the formula mass of Ca(ClO)2.
a. A The molar mass of S2Cl2 is obtained from its molecular mass as follows:
2S +2Cl S 2 Cl2
(2 atoms)(32.065 amu/atom) = (2 atoms)(35.453 amu/atom) = molecular mass of S 2 Cl2 =
64.130 amu 70.906 amu 135.036 amu
The molar mass of S2Cl2 is 135.036 g/mol. B The mass of 1.75 mol of S2Cl2 is calculated as follows:
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g moles S 2 Cl2 molar mass → mass of S 2 Cl2 (g) [ ( mol )] 135.036 g S Cl 2 2 1.75 mol S 2 Cl2 1 mol S 2 Cl2
= 236 g S Cl 2 2
b. A The formula mass of Ca(ClO)2 is obtained as follows:
1Ca 2Cl +2O Ca(ClO) 2
(1 atom)(40.078 amu/atom) (2 atoms)(35.453 amu/atom) (2 atoms)(15.9994 amu/atom) formula mass of Ca(ClO) 2
= = = =
40.078 amu 70.906 amu 31.9988 amu 142.983 amu
The molar mass of Ca(ClO)2 142.983 g/mol. B The mass of 1.75 mol of Ca(ClO)2 is calculated as follows:
molar mass Ca(ClO) 2 = mass Ca(ClO) 2 [ 1 mol Ca(ClO) 2 ] 142.983 g Ca(ClO) 2 1.75 mol Ca(ClO) 2 = 250 g Ca(ClO) 2 1 mol Ca(ClO) 2 moles Ca(ClO) 2
Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is. Exercise Calculate the mass of 0.0122 mol of each compound. a. Si3N4 (silicon nitride), used as bearings and rollers b. (CH3)3N (trimethylamine), a corrosion inhibitor Answer:
3.1 The Mole and Molar Masses
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a. 1.71 g b. 0.721 g
Summary The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12 and consists of Avogadro’s number (6.022 × 1023) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 1023 atoms, molecules, or formula units of that substance.
KEY TAKEAWAY • To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole.
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CONCEPTUAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Conceptual Problems. 1. Describe the relationship between an atomic mass unit and a gram. 2. Is it correct to say that ethanol has a formula mass of 46? Why or why not? 3. If 2 mol of sodium react completely with 1 mol of chlorine to produce sodium chloride, does this mean that 2 g of sodium reacts completely with 1 g of chlorine to give the same product? Explain your answer. 4. Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi3O8), a mineral used in the manufacture of porcelain. 5. Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass.
ANSWER 5. A = %N by mass, expressed as a decimal
1 molar mass of nitrogen in g ×A ×B g nitroglycerin ⎯→ gN ⎯→ mol N
B =
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems. 1. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass. 2. Calculate the molecular mass or formula mass of each compound. a. b. c. d. e. f. g. h.
KCl (potassium chloride) NaCN (sodium cyanide) H2S (hydrogen sulfide) NaN3 (sodium azide) H2CO3 (carbonic acid) K2O (potassium oxide) Al(NO3)3 (aluminum nitrate) Cu(ClO4)2 [copper(II) perchlorate]
3. Calculate the molecular mass or formula mass of each compound. a. b. c. d. e. f. g.
V2O4 (vanadium(IV) oxide) CaSiO3 (calcium silicate) BiOCl (bismuth oxychloride) CH3COOH (acetic acid) Ag2SO4 (silver sulfate) Na2CO3 (sodium carbonate) (CH3)2CHOH (isopropyl alcohol)
4. Calculate the molar mass of each compound. a.
b.
c.
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d.
e.
5. Calculate the molar mass of each compound. a.
b.
c.
d.
6. For each compound, write the condensed formula, name the compound, and give its molar mass. a.
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b.
7. For each compound, write the condensed formula, name the compound, and give its molar mass. a.
b.
8. Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or formula units are present in each sample? a. b. c. d. e.
CaO (lime) CaCO3 (chalk) C12H22O11 [sucrose (cane sugar)] NaOCl (bleach) CO2 (dry ice)
9. Calculate the mass in grams of each sample. a. 0.520 mol of N2O4 b. 1.63 mol of C6H4Br2 c. 4.62 mol of (NH4)2SO3 10. Give the number of molecules or formula units in each sample. a. 1.30 × 10−2 mol of SCl2 b. 1.03 mol of N2O5
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c. 0.265 mol of Ag2Cr2O7 11. Give the number of moles in each sample. a. 9.58 × 1026 molecules of Cl2 b. 3.62 × 1027 formula units of KCl c. 6.94 × 1028 formula units of Fe(OH)2 12. Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I2)? 13. What is the total number of atoms in each sample? a. b. c. d.
0.431 mol of Li 2.783 mol of methanol (CH3OH) 0.0361 mol of CoCO3 1.002 mol of SeBr2O
14. What is the total number of atoms in each sample? a. b. c. d.
0.980 mol of Na 2.35 mol of O2 1.83 mol of Ag2S 1.23 mol of propane (C3H8)
15. What is the total number of atoms in each sample? a. b. c. d.
2.48 g of HBr 4.77 g of CS2 1.89 g of NaOH 1.46 g of SrC2O4
16. Decide whether each statement is true or false and explain your reasoning. a. b. c. d.
There are more molecules in 0.5 mol of Cl2 than in 0.5 mol of H2. One mole of H2 has 6.022 × 1023 hydrogen atoms. The molecular mass of H2O is 18.0 amu. The formula mass of benzene is 78 amu.
17. Complete the following table. Mass Substance (g) MgCl2
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Number of Moles
Number of Molecules or Formula Units
Number of Atoms or Ions
37.62
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Mass Substance (g) AgNO3
Number of Moles
Number of Molecules or Formula Units
2.84 8.93 × 1025
BH4Cl
7.69 × 1026
K 2S H2SO4 C6H14 HClO3
Number of Atoms or Ions
1.29 11.84 2.45 × 1026
18. Give the formula mass or the molecular mass of each substance. a. b. c. d.
PbClF Cu2P2O7 BiONO3 Tl2SeO4
19. Give the formula mass or the molecular mass of each substance. a. b. c. d.
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MoCl5 B 2O3 UO2CO3 NH4UO2AsO4
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3.2 Determining Empirical and Molecular Formulas LEARNING OBJECTIVES 1. To determine the empirical formula of a compound from its composition by mass. 2. To derive the molecular formula of a compound from its empirical formula.
When a new chemical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, chemists determine its elemental composition, its empirical formula, and its structure to understand its properties. In this section, we focus on how to determine the empirical formula of a compound and then use it to determine the molecular formula if the molar mass of the compound is known.
Calculating Mass Percentages The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass; that is, the percent composition9—the percentage of each element present in a pure substance—is constant (although we now know there are exceptions to this law). For example, sucrose (cane sugar) is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. First we will use the molecular formula of sucrose (C 12H22O11) to calculate the mass percentage of the component elements; then we will show how mass percentages can be used to determine an empirical formula.
9. The percentage of each element present in a pure substance. With few exceptions, the percent composition of a chemical compound is constant (see law of definite proportions).
According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. We can use this information to calculate the mass of each element in 1 mol of sucrose, which will give us the molar mass of sucrose. We can then use these masses to calculate the percent composition of sucrose. To three decimal places, the calculations are the following:
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Equation 3.3
mass of C/mol of sucrose = 12 mol C × mass of H/mol of sucrose = 22 mol H × mass of O/mol of sucrose = 11 mol O ×
12.011 g C 1 mol C 1.008 g H 1 mol H 15.999 g O 1 mol O
= 144.132 g C = 22.176 g H = 175.989 g O
Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon. The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places:
144.132 g C mass of C/mol sucrose × 100 = molar mass of sucrose 342.297 g/mol 22.176 g H mass of H/mol sucrose mass % H in sucrose = × 100 = molar mass of sucrose 342.297 g/mol 175.989 g O mass of O/mol sucrose mass % O in sucrose = × 100 = molar mass of sucrose 342.297 g/mol mass % C in sucrose =
You can check your work by verifying that the sum of the percentages of all the elements in the compound is 100%: 42.12% + 6.48% + 51.41% = 100.01% If the sum is not 100%, you have made an error in your calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.
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We could also calculate the mass percentages using atomic masses and molecular masses, with atomic mass units. Because the answer we are seeking is a ratio, expressed as a percentage, the units of mass cancel whether they are grams (using molar masses) or atomic mass units (using atomic and molecular masses).
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EXAMPLE 5 Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C14H18N2O5.
a. Calculate the mass percentage of each element in aspartame. b. Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame. Given: molecular formula and mass of sample Asked for: mass percentage of all elements and mass of one element in sample Strategy: A Use atomic masses from the periodic table to calculate the molar mass of aspartame. B Divide the mass of each element by the molar mass of aspartame; then multiply by 100 to obtain percentages. C To find the mass of an element contained in a given mass of aspartame, multiply the mass of aspartame by the mass percentage of that element, expressed as a decimal.
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Solution:
a. A We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:
14C
(14 mol C)(12.011 g/mol C) =
168.154 g
2N
(2 mol N)(14.007 g/mol N) =
28.014 g
18H +5O
C 14 H 18 N 2 O 5
(18 mol H)(1.008 g/mol H) = (5 mol O)(15.999 g/mol O) = molar mass of aspartame =
18.114 g
79.995 g
294.277 g/mo
Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g). B To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:
168.154 g C 294.277 g aspartame 18.114 g H mass % H = 294.277 g aspartame 28.014 g N mass % N = 294.277 g aspartame 79.995 g O mass % O = 294.277 g aspartame mass % C =
× 100 = 57.14% C × 100 = 6.16% H × 100 = 9.52% N × 100 = 27.18% O
As a check, we can add the percentages together: 57.14% + 6.16% + 9.52% + 27.18% = 100.00% If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation.
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b. C The mass of carbon in 1.00 g of aspartame is calculated as follows:
mass of C = 1.00 g aspartame ×
57.14 g C 100 g aspartame
= 0.571 g C
Exercise Calculate the mass percentage of each element in aluminum oxide (Al 2O3). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide. Answer: 52.93% aluminum; 47.08% oxygen; 1.92 g Al
Determining the Empirical Formula of Penicillin Just as we can use the empirical formula of a substance to determine its percent composition, we can use the percent composition of a sample to determine its empirical formula, which can then be used to determine its molecular formula. Such a procedure was actually used to determine the empirical and molecular formulas of the first antibiotic to be discovered: penicillin. Antibiotics are chemical compounds that selectively kill microorganisms, many of which cause diseases. Although we may take antibiotics for granted today, penicillin was discovered only about 80 years ago. The subsequent development of a wide array of other antibiotics for treating many common diseases has contributed greatly to the substantial increase in life expectancy over the past 50 years. The discovery of penicillin is a historical detective story in which the use of mass percentages to determine empirical formulas played a key role. In 1928, Alexander Fleming, a young microbiologist at the University of London, was working with a common bacterium that causes boils and other infections such as blood poisoning. For laboratory study, bacteria are commonly grown on the surface of a nutrient-containing gel in small, flat culture dishes. One day Fleming noticed that one of his cultures was contaminated by a bluish-green mold similar to the mold found on spoiled bread or fruit. Such accidents are rather common, and most laboratory workers would have simply thrown the cultures away. Fleming noticed, however, that the bacteria were growing everywhere on the gel except near the contaminating mold (part (a) in Figure 3.3 " "), and he hypothesized that the
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mold must be producing a substance that either killed the bacteria or prevented their growth. To test this hypothesis, he grew the mold in a liquid and then filtered the liquid and added it to various bacteria cultures. The liquid killed not only the bacteria Fleming had originally been studying but also a wide range of other disease-causing bacteria. Because the mold was a member of the Penicillium family (named for their pencil-shaped branches under the microscope) (part (b) in Figure 3.3 " "), Fleming called the active ingredient in the broth penicillin. Figure 3.3 Penicillium
(a) Penicillium mold is growing in a culture dish; the photo shows its effect on bacterial growth. (b) In this photomicrograph of Penicillium, its rod- and pencil-shaped branches are visible. The name comes from the Latin penicillus, meaning “paintbrush.”
Although Fleming was unable to isolate penicillin in pure form, the medical importance of his discovery stimulated researchers in other laboratories. Finally, in 1940, two chemists at Oxford University, Howard Florey (1898–1968) and Ernst Chain (1906–1979), were able to isolate an active product, which they called penicillin G. Within three years, penicillin G was in widespread use for treating pneumonia, gangrene, gonorrhea, and other diseases, and its use greatly increased the survival rate of wounded soldiers in World War II. As a result of their work, Fleming, Florey, and Chain shared the Nobel Prize in Medicine in 1945. As soon as they had succeeded in isolating pure penicillin G, Florey and Chain subjected the compound to a procedure called combustion analysis (described later in this section) to determine what elements were present and in what quantities. The results of such analyses are usually reported as mass percentages. They
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discovered that a typical sample of penicillin G contains 53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur, and 6.5% sodium by mass. The sum of these numbers is only 82.1%, rather than 100.0%, which implies that there must be one or more additional elements. A reasonable candidate is oxygen, which is a common component of compounds that contain carbon and hydrogen;Do not assume that the “missing” mass is always due to oxygen. It could be any other element. for technical reasons, however, it is difficult to analyze for oxygen directly. If we assume that all the missing mass is due to oxygen, then penicillin G contains (100.0% − 82.1%) = 17.9% oxygen. From these mass percentages, the empirical formula and eventually the molecular formula of the compound can be determined. To determine the empirical formula from the mass percentages of the elements in a compound such as penicillin G, we need to convert the mass percentages to relative numbers of atoms. For convenience, we assume that we are dealing with a 100.0 g sample of the compound, even though the sizes of samples used for analyses are generally much smaller, usually in milligrams. This assumption simplifies the arithmetic because a 53.9% mass percentage of carbon corresponds to 53.9 g of carbon in a 100.0 g sample of penicillin G; likewise, 4.8% hydrogen corresponds to 4.8 g of hydrogen in 100.0 g of penicillin G; and so forth for the other elements. We can then divide each mass by the molar mass of the element to determine how many moles of each element are present in the 100.0 g sample:
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Equation 3.4
mass (g) mol = (g) = mol molar mass (g/mol) ( g ) 1 mol C 53.9 g C = 4.49 mol C 12.011 g C 1 mol H 4.8 g H = 4.8 mol H 1.008 g H 1 mol N 7.9 g N = 0.56 mol N 14.007 g N 1 mol S 9.0 g S = 0.28 mol S 32.065 g S 1 mol Na 6.5 g Na = 0.28 mol Na 22.990 g Na 1 mol O 17.9 g O = 1.12 mol O 15.999 g O
Thus 100.0 g of penicillin G contains 4.49 mol of carbon, 4.8 mol of hydrogen, 0.56 mol of nitrogen, 0.28 mol of sulfur, 0.28 mol of sodium, and 1.12 mol of oxygen (assuming that all the missing mass was oxygen). The number of significant figures in the numbers of moles of elements varies between two and three because some of the analytical data were reported to only two significant figures. These results tell us the ratios of the moles of the various elements in the sample (4.49 mol of carbon to 4.8 mol of hydrogen to 0.56 mol of nitrogen, and so forth), but they are not the whole-number ratios we need for the empirical formula—the empirical formula expresses the relative numbers of atoms in the smallest whole numbers possible. To obtain whole numbers, we divide the numbers of moles of all the elements in the sample by the number of moles of the element present in the lowest relative amount, which in this example is sulfur or sodium. The results will
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be the subscripts of the elements in the empirical formula. To two significant figures, the results are Equation 3.5
4.49 4.8 0.56 = 16 H: = 17 N: = 2.0 0.28 0.28 0.28 0.28 0.28 1.12 S: = 1.0 Na: = 1.0 O: = 4.0 0.28 0.28 0.28 C:
The empirical formula of penicillin G is therefore C16H17N2NaO4S. Other experiments have shown that penicillin G is actually an ionic compound that contains Na+ cations and [C16H17N2O4S]− anions in a 1:1 ratio. The complex structure of penicillin G (Figure 3.4 "Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G") was not determined until 1948. Figure 3.4 Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G
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In some cases, one or more of the subscripts in a formula calculated using this procedure may not be integers. Does this mean that the compound of interest contains a nonintegral number of atoms? No; rounding errors in the calculations as well as experimental errors in the data can result in nonintegral ratios. When this happens, you must exercise some judgment in interpreting the results, as illustrated in Example 6. In particular, ratios of 1.50, 1.33, or 1.25 suggest that you should multiply all subscripts in the formula by 2, 3, or 4, respectively. Only if the ratio is within 5% of an integral value should you consider rounding to the nearest integer.
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EXAMPLE 6 Calculate the empirical formula of the ionic compound calcium phosphate, a major component of fertilizer and a polishing agent in toothpastes. Elemental analysis indicates that it contains 38.77% calcium, 19.97% phosphorus, and 41.27% oxygen. Given: percent composition Asked for: empirical formula Strategy: A Assume a 100 g sample and calculate the number of moles of each element in that sample. B Obtain the relative numbers of atoms of each element in the compound by dividing the number of moles of each element in the 100 g sample by the number of moles of the element present in the smallest amount. C If the ratios are not integers, multiply all subscripts by the same number to give integral values. D Because this is an ionic compound, identify the anion and cation and write the formula so that the charges balance. Solution: A A 100 g sample of calcium phosphate contains 38.77 g of calcium, 19.97 g of phosphorus, and 41.27 g of oxygen. Dividing the mass of each element in the 100 g sample by its molar mass gives the number of moles of each element in the sample:
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moles Ca = 38.77 g Ca × moles P = 19.97 g P × moles O = 41.27 g O ×
1 mol Ca 40.078 g Ca 1 mol P
30.9738 g P 1 mol O 15.9994 g O
= 0.9674 mol Ca
= 0.6447 mol P = 2.5800 mol O
B To obtain the relative numbers of atoms of each element in the compound, divide the number of moles of each element in the 100-g sample by the number of moles of the element in the smallest amount, in this case phosphorus:
P:
0.6447 mol P 0.9674 2.5800 = 1.000 Ca: = 1.501 O: = 4.002 0.6447 mol P 0.6447 0.6447 C We could write the empirical formula of calcium phosphate as Ca1.501P1.000O4.002, but the empirical formula should show the ratios of the elements as small whole numbers. To convert the result to integral form, multiply all the subscripts by 2 to get Ca3.002P2.000O8.004. The deviation from integral atomic ratios is small and can be attributed to minor experimental errors; therefore, the empirical formula is Ca3P2O8. D The calcium ion (Ca2+) is a cation, so to maintain electrical neutrality, phosphorus and oxygen must form a polyatomic anion. We know from Chapter 2 "Molecules, Ions, and Chemical Formulas" that phosphorus and oxygen form the phosphate ion (PO43−; see Table 2.4 "Common Polyatomic Ions and Their Names"). Because there are two phosphorus atoms in the empirical formula, two phosphate ions must be present. So we write the formula of calcium phosphate as Ca3(PO4)2. Exercise Calculate the empirical formula of ammonium nitrate, an ionic compound that contains 35.00% nitrogen, 5.04% hydrogen, and 59.96% oxygen by mass; refer to Table 2.4 "Common Polyatomic Ions and Their Names" if necessary. Although ammonium nitrate is widely used as a fertilizer, it can be dangerously explosive. For example, it was a major component of the explosive used in the 1995 Oklahoma City bombing.
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Answer: N2H4O3 is NH4+NO3−, written as NH4NO3
Combustion Analysis One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure 3.5 "Steps for Obtaining an Empirical Formula from Combustion Analysis", and a typical combustion analysis is illustrated in Example 7. Figure 3.5 Steps for Obtaining an Empirical Formula from Combustion Analysis
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EXAMPLE 7 Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO 2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene. Given: mass of sample and mass of combustion products Asked for: empirical formula Strategy: A Use the masses and molar masses of the combustion products, CO 2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene. B Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene. Solution: A Upon combustion, 1 mol of CO2 is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO 2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
mass of C = 69.00 mg CO2 ×
1 g 1000 mg
×
1 mol CO 2 44.010 g CO 2
×
1 m
1 mo
= 1.883 × 10 −2 g C mass of H = 11.30 mg H2 O ×
1 g 1000 mg
×
1 mol H2 O 18.015 g H2 O
×
= 1.264 × 10 −3 g H
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1 mo
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B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
moles C = 1.883 × 10 –2 g C × moles H = 1.264 × 10 –3 g H ×
1 mol C 12.011 g C 1 mol H 1.0079 g H
= 1.568 × 10 –3 mol C = 1.254 × 10 –3 mol H
Dividing each number by the number of moles of the element present in the smaller amount gives
H:
1.254 × 10 −3 1.254 × 10 −3
= 1.000 C:
1.568 × 10 −3 1.254 × 10 −3
= 1.250
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results. Exercise a. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene. b. The empirical formula of benzene is CH (its molecular formula is C 6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced? Answer: a. The empirical formula is C4H5. (The molecular formula of xylene is actually C8H10.) b. 33.81 mg of CO2; 6.92 mg of H2O
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From Empirical Formula to Molecular Formula The empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio. For a covalent substance, we are usually more interested in the molecular formula, which gives the actual number of atoms of each kind present per molecule. Without additional information, however, it is impossible to know whether the formula of penicillin G, for example, is C 16H17N2NaO4S or an integral multiple, such as C32H34N4Na2O8S2, C48H51N6Na3O12S3, or (C16H17N2NaO4S)n, where n is an integer. (The actual structure of penicillin G is shown in Figure 3.4 "Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G".) Consider glucose, the sugar that circulates in our blood to provide fuel for our bodies and especially for our brains. Results from combustion analysis of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Because combustion occurs in the presence of oxygen, it is impossible to directly determine the percentage of oxygen in a compound by using combustion analysis; other more complex methods are necessary. If we assume that the remaining percentage is due to oxygen, then glucose would contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of oxygen. To calculate the number of moles of each element in the 100.0 g sample, we divide the mass of each element by its molar mass: Equation 3.6
moles C = 39.68 g C × moles H = 6.58 g H × moles O = 53.79 g O ×
1 mol C 12.011 g C 1 mol H 1.0079 g H 1 mol O 15.9994 g O
= 3.304 mol C = 6.53 mol H = 3.362 mol O
Once again, we find the subscripts of the elements in the empirical formula by dividing the number of moles of each element by the number of moles of the element present in the smallest amount:
C:
3.304 6.53 3.362 = 1.000 H: = 1.98 O: = 1.018 3.304 3.304 3.304
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The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is approximately 2. The empirical formula of glucose is therefore CH 2O, but what is its molecular formula? Many known compounds have the empirical formula CH2O, including formaldehyde, which is used to preserve biological specimens and has properties that are very different from the sugar circulating in our blood. At this point, we cannot know whether glucose is CH2O, C2H4O2, or any other (CH2O)n. We can, however, use the experimentally determined molar mass of glucose (180 g/mol) to resolve this dilemma. First, we calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose, Equation 3.7
formula mass of CH 2 O =
[
1 mol C
( 1 mol C )] 12.011 g
+
[
2 mol H
This is much smaller than the observed molar mass of 180 g/mol. Second, we determine the number of formula units per mole. For glucose, we can calculate the number of (CH2O) units—that is, the n in (CH2O)n—by dividing the molar mass of glucose by the formula mass of CH2O: Equation 3.8
n=
180 g = 5.99 ≈ 6 CH2 O formula units 30.026 g/CH 2 O
Each glucose contains six CH2O formula units, which gives a molecular formula for glucose of (CH2O)6, which is more commonly written as C6H12O6. The molecular structures of formaldehyde and glucose, both of which have the empirical formula CH2O, are shown in Figure 3.6 "Structural Formulas and Ball-and-Stick Models of (a) Formaldehyde and (b) Glucose".
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(1
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Figure 3.6 Structural Formulas and Ball-and-Stick Models of (a) Formaldehyde and (b) Glucose
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EXAMPLE 8 Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol. Given: percent composition and molar mass Asked for: molecular formula Strategy: A Assume 100 g of caffeine. From the percentages given, use the procedure given in Example 6 to calculate the empirical formula of caffeine. B Calculate the formula mass and then divide the experimentally determined molar mass by the formula mass. This gives the number of formula units present. C Multiply each subscript in the empirical formula by the number of formula units to give the molecular formula. Solution: A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine.
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moles C = 49.18 g C × moles H = 5.39 g H × moles N = 28.65 g N × moles O = 16.68 g O ×
1 mol C 12.011 g C 1 mol H 1.0079 g H 1 mol N 14.0067 g N 1 mol O 15.9994 g O
= 4.095 mol C = 5.35 mol H = 2.045 mol N = 1.043 mol O
To obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount:
O:
1.043 4.095 5.35 2.045 = 1.000 C: = 3.926 H: = 5.13 N: 1.043 1.043 1.043 1.043
These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C 4H5N2O. B The molecular formula of caffeine could be C4H5N2O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows:
4C
(4 atoms C)(12.011 g/atom C) =
48.044 g
2N
(2 atoms N)(14.0067 g/atom N) =
28.0134 g
C4 H5 N2 O
formula mass of caffeine =
97.096 g
5H +1O
(5 atoms H)(1.0079 g/atom H) =
(1 atom O)(15.9994 g/atom O) =
5.0395 g
15.9994 g
Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives
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196 g/mol = 2.02 ≈ 2 C4 H5 N2 O empirical formula units 97.096 g/C 4 H5 N2 O C There are two C4H5N2O formula units in caffeine, so the molecular formula must be (C4H5N2O)2 = C8H10N4O2. The structure of caffeine is as follows:
Exercise Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer. Answer: C2Cl2F4
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Summary The empirical formula of a substance can be calculated from the experimentally determined percent composition, the percentage of each element present in a pure substance by mass. In many cases, these percentages can be determined by combustion analysis. If the molar mass of the compound is known, the molecular formula can be determined from the empirical formula.
KEY TAKEAWAY • The empirical formula of a substance can be calculated from its percent composition, and the molecular formula can be determined from the empirical formula and the compound’s molar mass.
CONCEPTUAL PROBLEMS 1. What is the relationship between an empirical formula and a molecular formula? 2. Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems. 1. What is the mass percentage of water in each hydrate? a. H3AsO4·0·5H2O b. NH4NiCl3·6H2O c. Al(NO3)3·9H2O 2. What is the mass percentage of water in each hydrate? a. CaSO4·2H2O b. Fe(NO3)3·9H2O c. (NH4)3ZrOH(CO3)3·2H2O 3. Which of the following has the greatest mass percentage of oxygen—KMnO 4, K2Cr2O7, or Fe2O3? 4. Which of the following has the greatest mass percentage of oxygen—ThOCl 2, MgCO3, or NO2Cl? 5. Calculate the percent composition of the element shown in bold in each compound. a. b. c. d.
SbBr3 As2I4 AlPO4 C6H10O
6. Calculate the percent composition of the element shown in bold in each compound. a. b. c. d.
HBrO3 CsReO4 C3H 8O FeSO4
7. A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound?
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8. The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds. Compound % Iron % Oxygen Empirical Formula 1
69.9
30.1
2
77.7
22.3
3
72.4
27.6
9. What is the mass percentage of water in each hydrate? a. LiCl·H2O b. MgSO4·7H2O c. Sr(NO3)2·4H2O 10. What is the mass percentage of water in each hydrate? a. CaHPO4·2H2O b. FeCl2·4H2O c. Mg(NO3)2·4H2O 11. Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates? Compound Initial Mass (g) Mass after Cooling (g) NiSO4·xH2O
2.08
1.22
CoCl2·xH2O
1.62
0.88
12. Which contains the greatest mass percentage of sulfur—FeS2, Na2S2O4, or Na2S? 13. Given equal masses of each, which contains the greatest mass percentage of sulfur—NaHSO4 or K2SO4? 14. Calculate the mass percentage of oxygen in each polyatomic ion. a. b. c. d.
bicarbonate chromate acetate sulfite
15. Calculate the mass percentage of oxygen in each polyatomic ion.
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a. b. c. d.
oxalate nitrite dihydrogen phosphate thiocyanate
16. The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula? 17. A compound has the empirical formula C2H4O, and its formula mass is 88 g. What is its molecular formula? 18. Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula? 19. How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene?
20. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline?
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21. How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame?
22. How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G?
23. Combustion of a 34.8 mg sample of benzaldehyde, which contains only carbon, hydrogen, and oxygen, produced 101 mg of CO2 and 17.7 mg of H2O. a. What was the mass of carbon and hydrogen in the sample?
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b. Assuming that the original sample contained only carbon, hydrogen, and oxygen, what was the mass of oxygen in the sample? c. What was the mass percentage of oxygen in the sample? d. What is the empirical formula of benzaldehyde? e. The molar mass of benzaldehyde is 106.12 g/mol. What is its molecular formula? 24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O. a. b. c. d.
What is the mass of oxygen in the sample? What is the mass percentage of oxygen in the sample? What is the empirical formula of salicylic acid? The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula?
25. Given equal masses of the following acids, which contains the greatest amount of hydrogen that can dissociate to form H+—nitric acid, hydroiodic acid, hydrocyanic acid, or chloric acid? 26. Calculate the formula mass or the molecular mass of each compound. a. b. c. d. e. f.
heptanoic acid (a seven-carbon carboxylic acid) 2-propanol (a three-carbon alcohol) KMnO4 tetraethyllead sulfurous acid ethylbenzene (an eight-carbon aromatic hydrocarbon)
27. Calculate the formula mass or the molecular mass of each compound. a. b. c. d. e. f.
MoCl5 B 2O3 bromobenzene cyclohexene phosphoric acid ethylamine
28. Given equal masses of butane, cyclobutane, and propene, which contains the greatest mass of carbon? 29. Given equal masses of urea [(NH2)2CO] and ammonium sulfate, which contains the most nitrogen for use as a fertilizer?
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ANSWERS 1. To two decimal places, the percentages are: a. 5.97% b. 37.12% c. 43.22% 3. % oxygen: KMnO4, 40.50%; K2Cr2O7, 38.07%; Fe2O3, 30.06% 5. To two decimal places, the percentages are: a. b. c. d.
66.32% Br 22.79% As 25.40% P 73.43% C
7. Cr2O3. 9. To two decimal places, the percentages are: a. 29.82% b. 51.16% c. 25.40% 11. NiSO4 · 6H2O and CoCl2 · 6H2O 13. NaHSO4 15. a. b. c. d.
72.71% 69.55% 65.99% 0%
17. C4H8O2 23. a. b. c. d. e.
27.6 mg C and 1.98 mg H 5.2 mg O 15% C7H6O C7H6O
25. hydrocyanic acid, HCN 27. To two decimal places, the values are:
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a. b. c. d. e. f.
273.23 amu 69.62 amu 157.01 amu 82.14 amu 98.00 amu 45.08 amu
29. Urea
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3.3 Chemical Equations LEARNING OBJECTIVES 1. To describe a chemical reaction. 2. To calculate the quantities of compounds produced or consumed in a chemical reaction.
As shown in Figure 3.7 "An Ammonium Dichromate Volcano: Change during a Chemical Reaction", applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. We can describe this reaction with a chemical equation10, an expression that gives the identities and quantities of the substances in a chemical reaction. Chemical formulas and other symbols are used to indicate the starting material(s), or reactant(s)11, which by convention are written on the left side of the equation, and the final compound(s), or product(s)12, which are written on the right. An arrow points from the reactant to the products: Figure 3.7 An Ammonium Dichromate Volcano: Change during a Chemical Reaction
10. An expression that gives the identities and quantities of the substances in a chemical reaction. Chemical formulas are used to indicate the reactants on the left and the products on the right. An arrow points from reactants to products. 11. The starting material(s) in a chemical reaction. 12. The final compound(s) produced in a chemical reaction.
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The starting material (left) is solid ammonium dichromate. A chemical reaction (right) transforms it to solid chromium(III) oxide, depicted showing a portion of its chained structure, nitrogen gas, and water vapor. (In addition, energy in the form of heat and light is released.) During the reaction, the distribution of atoms changes, but the number of atoms of each element does not change. Because the numbers of each type of atom are the same in the reactants and the products, the chemical equation is balanced.
Equation 3.9
(NH4 )2 Cr2 O7 → Cr2 O3 + N2 + 4H2 O reactant
products
The arrow is read as “yields” or “reacts to form.” So Equation 3.9 tells us that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products). The equation for this reaction is even more informative when written as Equation 3.10 (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) Equation 3.10 is identical to Equation 3.9 except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water. Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of Equation 3.9 and Equation 3.10. (For more information on the law of conservation of mass, see Section 1.4 "A Brief History of Chemistry".) As illustrated in Figure 3.7 "An Ammonium Dichromate Volcano: Change during a Chemical Reaction", each side has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms. In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. Equation 3.9 and Equation 3.10 are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr–O and N–H bonds), and new bonds are formed to create the products (here, O–H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced,
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and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced.
Note the Pattern A chemical reaction changes only the distribution of atoms, not the number of atoms.
Interpreting Chemical Equations In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced chemical equation provides quantitative information. Specifically, it tells the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced chemical equation is the coefficient13 of that species (e.g., the 4 preceding H2O in Equation 3.9). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure 3.8 "The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced Chemical Reaction for the Ammonium Dichromate Volcano", the coefficients allow us to interpret Equation 3.9 in any of the following ways: • Two NH4+ ions and one Cr2O72− ion yield 1 formula unit of Cr2O3, 1 N2 molecule, and 4 H2O molecules. • One mole of (NH4)2Cr2O7 yields 1 mol of Cr2O3, 1 mol of N2, and 4 mol of H2O. • A mass of 252 g of (NH4)2Cr2O7 yields 152 g of Cr2O3, 28 g of N2, and 72 g of H2O. • A total of 6.022 × 1023 formula units of (NH4)2Cr2O7 yields 6.022 × 1023 formula units of Cr2O3, 6.022 × 1023 molecules of N2, and 24.09 × 1023 molecules of H2O. 13. A number greater than 1 preceding a formula in a balanced chemical equation and indicating the number of atoms, molecules, or formula units of a reactant or a product.
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Figure 3.8 The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced Chemical Reaction for the Ammonium Dichromate Volcano
These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio14. For example, the mole ratio of H2O to N2 in Equation 3.9 is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass: 252 g of (NH4)2Cr2O7 yields 152 + 28 + 72 = 252 g of products. The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. We will consider these issues in more detail in later chapters.
14. The ratio of the number of moles of one substance to the number of moles of another, as depicted by a balanced chemical equation.
3.3 Chemical Equations
An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose (C 6H12O6), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.
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EXAMPLE 9 The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Construct a table showing how to interpret the information in this equation in terms of a. b. c. d.
a single molecule of glucose. moles of reactants and products. grams of reactants and products represented by 1 mol of glucose. numbers of molecules of reactants and products represented by 1 mol of glucose.
Given: balanced chemical equation Asked for: molecule, mole, and mass relationships Strategy: A Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios. B Use the molar masses of the reactants and products to convert from moles to grams. C Use Avogadro’s number to convert from moles to the number of molecules. Solution: This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information. a. A One molecule of glucose reacts with 6 molecules of O2 to yield 6 molecules of CO2 and 6 molecules of H2O. b. One mole of glucose reacts with 6 mol of O2 to yield 6 mol of CO2 and 6 mol of H2O.
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c. B To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O2, 31.9988; CO2, 44.010; and H2O, 18.015.
180.16 g 1 mol glucose 1 mol glucose
mass of reactants = m
+ 6 mol O 2
31.9988 g 1 mol O 2
=6
g glucose + g O 2 = g
372.15 g = 37 d. C One mole of glucose contains Avogadro’s number (6.022 × 1023) of glucose molecules. Thus 6.022 × 1023 glucose molecules react with (6 × 6.022 × 1023) = 3.613 × 1024 oxygen molecules to yield (6 × 6.022 × 1023) = 3.613 × 1024 molecules each of CO2 and H2O. In tabular form: C6H12O6(s) +
6O2(g)
→
6CO2(g)
+
6H2O(l)
a. 1 molecule
6 molecules
6 molecules
6 molecules
b. 1 mol
6 mol
6 mol
6 mol
c. 180.16 g
191.9928 g
264.06 g
108.09 g
3.613 × 1024 molecules
3.613 × 1024 molecules
3.613 × 1024 molecules
d.
6.022 × 1023 molecules
Exercise Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas. The explosion resulted from the following reaction: 2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
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Construct a table showing how to interpret the information in the equation in terms of a. b. c. d.
individual molecules and ions. moles of reactants and products. grams of reactants and products given 2 mol of ammonium nitrate. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
Answer: 2NH4NO3(s) a.
2NH4+ ions and 2NO3− ions
→
2N2(g)
+
4H2O(g)
+
O2(g)
2 molecules
4 molecules
1 molecule
b. 2 mol
2 mol
4 mol
1 mol
c. 160.0864 g
56.0268 g
72.0608 g
31.9988 g
1.204 × 1024 molecules
2.409 × 1024 molecules
6.022 × 1023 molecules
d.
1.204 × 1024 formula units
Ammonium nitrate can be hazardous. This aerial photograph of Texas City, Texas, shows the devastation caused by the explosion of a shipload
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of ammonium nitrate on April 16, 1947.
Balancing Simple Chemical Equations When a chemist encounters a new reaction, it does not usually come with a label that shows the balanced chemical equation. Instead, the chemist must identify the reactants and products and then write them in the form of a chemical equation that may or may not be balanced as first written. Consider, for example, the combustion of n-heptane (C7H16), an important component of gasoline: Equation 3.11 C7H16(l) + O2(g) → CO2(g) + H2O(g) The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water (Figure 3.9 "An Example of a Combustion Reaction"). Equation 3.11 is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, we must adjust the coefficients of the reactants and products to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, we cannot balance the equation by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure 3.10 "Balancing Equations".
3.3 Chemical Equations
Figure 3.9 An Example of a Combustion Reaction
The wax in a candle is a highmolecular-mass hydrocarbon, which produces gaseous carbon dioxide and water vapor in a combustion reaction. When the candle is allowed to burn inside a flask, drops of water, one of the products of combustion,
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Figure 3.10 Balancing Equations condense at the top of the inner surface of the flask.
You cannot change subscripts in a chemical formula to balance a chemical equation; you can change only the coefficients. Changing subscripts changes the ratios of atoms in the molecule and the resulting chemical properties. For example, water (H2O) and hydrogen peroxide (H2O2) are chemically distinct substances. H2O2 decomposes to H2O and O2 gas when it comes in contact with the metal platinum, whereas no such reaction occurs between water and platinum.
The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. We present an efficient approach to balancing a chemical equation using this method.
Steps in Balancing a Chemical Equation 1. Identify the most complex substance. 2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides. 3. Balance polyatomic ions (if present) as a unit. 4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional
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coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients. 5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced. To demonstrate this approach, let’s use the combustion of n-heptane (Equation 3.11) as an example. 1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is C 7H16. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance. 2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side: Equation 3.12 C7H16 + O2 → 7CO2 + H2O 3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction. 4. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: Equation 3.13 C7H16 + O2 → 7CO2 + 8H2O The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side: Equation 3.14 C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g) 5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.
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The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. Consider, for example, a similar reaction, the combustion of isooctane (C8H18). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows: Equation 3.15 C8H18(l) + O2(g) → CO2(g) + H2O(g) 1. Identify the most complex substance. Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane. 2. Adjust the coefficients. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products: Equation 3.16 C8H18 + O2 → 8CO2 + H2O 3. Balance polyatomic ions as a unit. This step does not apply to this equation. 4. Balance the remaining atoms. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products: Equation 3.17 C8H18 + O2 → 8CO2 + 9H2O The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms: Equation 3.18 C8H18 + 25/2O2 → 8CO2 + 9H2O Equation 3.18 is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2: Equation 3.19 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)
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5. Check your work. The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side. Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.
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EXAMPLE 10 The reaction of the mineral hydroxyapatite [Ca5(PO4)3(OH)] with phosphoric acid and water gives Ca(H2PO4)2·H2O (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction. Given: reactants and product Asked for: balanced chemical equation Strategy: A Identify the product and the reactants and then write the unbalanced chemical equation. B Follow the steps for balancing a chemical equation. Solution: A We must first identify the product and reactants and write an equation for the reaction. The formulas for hydroxyapatite and calcium dihydrogen phosphate monohydrate are given in the problem. Recall from Chapter 2 "Molecules, Ions, and Chemical Formulas" that phosphoric acid is H3PO4. The initial (unbalanced) equation is as follows: Ca5(PO4)3(OH)(s) + H3PO4(aq) + H2O(l) → Ca(H2PO4)2·H2O(s) 1. B Identify the most complex substance. We start by assuming that only one molecule or formula unit of the most complex substance, Ca5(PO4)3(OH), appears in the balanced chemical equation. 2. Adjust the coefficients. Because calcium is present in only one reactant and one product, we begin with it. One formula unit of Ca5(PO4)3(OH) contains 5 calcium atoms, so we need 5 Ca(H2PO4)2·H2O on the right side: Ca5(PO4)3(OH) + H3PO4 + H2O → 5Ca(H2PO4)2·H2O 3. Balance polyatomic ions as a unit. It is usually easier to balance an equation if we recognize that certain combinations of
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atoms occur on both sides. In this equation, the polyatomic phosphate ion (PO43−), shows up in three places.In H3PO4, the phosphate ion is combined with three H+ ions to make phosphoric acid (H3PO4), whereas in Ca(H2PO4)2·H2O it is combined with two H+ ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO4 units on the right side but only 4 on the left. The simplest way to balance the PO4 units is to place a coefficient of 7 in front of H3PO4: Ca5(PO4)3(OH) + 7H3PO4 + H2O → 5Ca(H2PO4)2·H2O Although OH− is also a polyatomic ion, it does not appear on both sides of the equation. So oxygen and hydrogen must be balanced separately. 4. Balance the remaining atoms. We now have 30 hydrogen atoms on the right side but only 24 on the left. We can balance the hydrogen atoms using the least complex substance, H2O, by placing a coefficient of 4 in front of H2O on the left side, giving a total of 4 H2O molecules: Ca5(PO4)3(OH)(s) + 7H3PO4(aq) + 4H2O(l) → 5Ca(H2PO4)2·H2O(s) The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 45 oxygen atoms on each side. 5. Check your work. Both sides of the equation contain 5 calcium atoms, 7 phosphorus atoms, 30 hydrogen atoms, and 45 oxygen atoms. Exercise Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose are converted to ethanol and carbon dioxide. Write a balanced chemical reaction for the fermentation of glucose.
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Commercial use of fermentation. (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible.
Answer: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
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Summary In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right, separated by an arrow. In a balanced chemical equation, the numbers of atoms of each element and the total charge are the same on both sides of the equation. The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species. The mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation.
KEY TAKEAWAY • A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products.
CONCEPTUAL PROBLEMS 1. How does a balanced chemical equation agree with the law of definite proportions? 2. What is the difference between S8 and 8S? Use this example to explain why subscripts in a formula must not be changed. 3. What factors determine whether a chemical equation is balanced? 4. What information can be obtained from a balanced chemical equation? Does a balanced chemical equation give information about the rate of a reaction?
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NUMERICAL PROBLEMS 1. Balance each chemical equation. a. b. c. d. e. f.
KI(aq) + Br2(l) → KBr(aq) + I2(s) MnO2(s) + HCl(aq) → MnCl2(aq) + Cl2(g) + H2O(l) Na2O(s) + H2O(l) → NaOH(aq) Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s) SO2(g) + H2O(l) → H2SO3(aq) S2Cl2(l) + NH3(l) → S4N4(s) + S8(s) + NH4Cl(s)
2. Balance each chemical equation. a. b. c. d. e. f.
Be(s) + O2(g) → BeO(s) N2O3(g) + H2O(l) → HNO2(aq) Na(s) + H2O(l) → NaOH(aq) + H2(g) CaO(s) + HCl(aq) → CaCl2(aq) + H2O(l) CH3NH2(g) + O2(g) → H2O(g) + CO2(g) + N2(g) Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
3. Balance each chemical equation. a. b. c. d. e. f.
N2O5(g) → NO2(g) + O2(g) NaNO3(s) → NaNO2(s) + O2(g) Al(s) + NH4NO3(s) → N2(g) + H2O(l) + Al2O3(s) C3H5N3O9(l) → CO2(g) + N2(g) + H2O(g) + O2(g) reaction of butane with excess oxygen IO2F(s) + BrF3(l) → IF5(l) + Br2(l) + O2(g)
4. Balance each chemical equation. a. b. c. d. e. f.
H2S(g) + O2(g) → H2O(l) + S8(s) KCl(aq) + HNO3(aq) + O2(g) → KNO3(aq) + Cl2(g) + H2O(l) NH3(g) + O2(g) → NO(g) + H2O(g) CH4(g) + O2(g) → CO(g) + H2(g) NaF(aq) + Th(NO3)4(aq) → NaNO3(aq) + ThF4(s) Ca5(PO4)3F(s) + H2SO4(aq) + H2O(l) → H3PO4(aq) + CaSO4·2H2O(s) + HF(aq)
5. Balance each chemical equation. a. NaCl(aq) + H2SO4(aq) → Na2SO4(aq) + HCl(g) b. K(s) + H2O(l) → KOH(aq) + H2(g) c. reaction of octane with excess oxygen
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d. S8(s) + Cl2(g) → S2Cl2(l) e. CH3OH(l) + I2(s) + P4(s) → CH3I(l) + H3PO4(aq) + H2O(l) f. (CH3)3Al(s) + H2O(l) → CH4(g) + Al(OH)3(s) 6. Write a balanced chemical equation for each reaction. a. Aluminum reacts with bromine. b. Sodium reacts with chlorine. c. Aluminum hydroxide and acetic acid react to produce aluminum acetate and water. d. Ammonia and oxygen react to produce nitrogen monoxide and water. e. Nitrogen and hydrogen react at elevated temperature and pressure to produce ammonia. f. An aqueous solution of barium chloride reacts with a solution of sodium sulfate. 7. Write a balanced chemical equation for each reaction. a. b. c. d.
Magnesium burns in oxygen. Carbon dioxide and sodium oxide react to produce sodium carbonate. Aluminum reacts with hydrochloric acid. An aqueous solution of silver nitrate reacts with a solution of potassium chloride. e. Methane burns in oxygen. f. Sodium nitrate and sulfuric acid react to produce sodium sulfate and nitric acid.
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3.4 Mass Relationships in Chemical Equations LEARNING OBJECTIVE 1. To calculate the quantities of compounds produced or consumed in a chemical reaction.
A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. Stoichiometry15 is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A stoichiometric quantity16 is the amount of product or reactant specified by the coefficients in a balanced chemical equation. In Section 3.3 "Chemical Equations", for example, you learned how to express the stoichiometry of the reaction for the ammonium dichromate volcano in terms of the atoms, ions, or molecules involved and the numbers of moles, grams, and formula units of each (recognizing, for instance, that 1 mol of ammonium dichromate produces 4 mol of water). This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?
15. A collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. 16. The amount of product or reactant specified by the coefficients in a balanced chemical equation.
All these questions can be answered using the concepts of the mole and molar and formula masses, along with the coefficients in the appropriate balanced chemical equation.
Stoichiometry Problems When we carry out a reaction in either an industrial setting or a laboratory, it is easier to work with masses of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in Figure 3.11 "A Flowchart for Stoichiometric Calculations Involving Pure Substances" and described in the following text.
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Steps in Converting between Masses of Reactant and Product 1. Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass. 2. From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients). 3. Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess. Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons). Figure 3.11 A Flowchart for Stoichiometric Calculations Involving Pure Substances
The molar masses of the reactants and the products are used as conversion factors so that you can calculate the mass of product from the mass of reactant and vice versa.
To illustrate this procedure, let’s return to the combustion of glucose. We saw earlier that glucose reacts with oxygen to produce carbon dioxide and water: Equation 3.20 C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain doesn’t run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam,
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how many grams of carbon dioxide will you produce and exhale into the exam room? The initial step in solving a problem of this type must be to write the balanced chemical equation for the reaction. Inspection of Equation 3.20 shows that it is balanced as written, so we can proceed to the strategy outlined in Figure 3.11 "A Flowchart for Stoichiometric Calculations Involving Pure Substances", adapting it as follows: 1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:
moles glucose = 45.3 g glucose ×
1 mol glucose 180.2 g glucose
= 0.251 mol glu
2. According to the balanced chemical equation, 6 mol of CO 2 is produced per mole of glucose; the mole ratio of CO2 to glucose is therefore 6:1. The number of moles of CO2 produced is thus
6 mol CO2 1 mol glucose 6 mol CO2 = 0.251 mol glucose × 1 mol glucose
moles CO2 = mol glucose ×
= 1.51 mol CO2 3. Use the molar mass of CO2 (44.010 g/mol) to calculate the mass of CO2 corresponding to 1.51 mol of CO2:
mass of CO2 = 1.51 mol CO2 ×
44.010 g CO2 1 mol CO2
= 66.5 g CO2
We can summarize these operations as follows:
45.3 g glucose ×
44.010 g CO 2 1 mol glucose 6 mol CO 2 × × = 180.2 g glucose 1 mol glucose 1 mol CO 2 step 1
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step 2
step 3
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Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (For more information about rounding and significant digits, see Essential Skills 1 in Chapter 1 "Introduction to Chemistry", Section 1.9 "Essential Skills 1".) In Chapter 10 "Gases", you will discover that this amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. We could use similar methods to calculate the amount of oxygen consumed or the amount of water produced. We just used the balanced chemical equation to calculate the mass of product that is formed from a certain amount of reactant. We can also use the balanced chemical equation to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example 11, the mass of one reactant that is required to consume a given mass of another reactant.
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EXAMPLE 11 The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).
The US space shuttle Discovery during liftoff. The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine.
Given: reactants, products, and mass of one reactant Asked for: mass of other reactant Strategy:
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A Write the balanced chemical equation for the reaction. B Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons. Solution: We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons. A We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows: H2(g) + O2(g) → H2O(g) This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H 2O and H2 gives the balanced chemical equation: 2H2(g) + O2(g) → 2H2O(g) Thus 2 mol of H2 react with 1 mol of O2 to produce 2 mol of H2O.
1. B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:
mass of O 2 = 1.00 tn ×
2000 lb tn
×
453.6 g lb
= 9.07 × 10 5 g O 2
Using the molar mass of O2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O2 contained in this mass of O2:
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mol O 2 = 9.07 × 10 5 g O 2 ×
1 mol O 2 32.00 g O 2
= 2.83 × 10 4 mol O
2. Now use the coefficients in the balanced chemical equation to obtain the number of moles of H2 needed to react with this number of moles of O2:
mol H 2 = mol O 2 ×
2 mol H 2 1 mol O 2
= 2.83 × 10 4 mol O 2 ×
2 mol H 2 1 mol O 2
= 5.66 × 10 4 mol H
3. The molar mass of H2 (2.016 g/mol) allows us to calculate the corresponding mass of H2:
mass of H 2 = 5.66 × 10 4 mol H 2 ×
2.016 g H 2 mol H 2
= 1.14 × 10 5 g
Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors:
tons H 2 = 1.14 × 10 5 g H 2 ×
1 lb 453.6 g
×
1 tn 2000 lb
= 0.126 tn
The space shuttle had to be designed to carry 0.126 tn of H 2 for each 1.00 tn of O2. Even though 2 mol of H2 are needed to react with each mole of O2, the molar mass of H2 is so much smaller than that of O2 that only a relatively small mass of H2 is needed compared to the mass of O2. Exercise Alchemists produced elemental mercury by roasting the mercurycontaining ore cinnabar (HgS) in air:
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HgS(s) + O2(g) → Hg(l) + SO2(g) The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction? Answer: 86.2 g
Limiting Reactants In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant17. The reactant that remains after a reaction has gone to completion is in excess. To be certain you understand these concepts, let’s first consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus Equation 3.21 1 box mix + 2 eggs → 1 batch brownies
17. The reactant that restricts the amount of product obtained in a chemical reaction.
If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in Equation 3.21 is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant (Figure 3.12 "The Concept of a Limiting Reactant in the Preparation of Brownies"). Even if you had a refrigerator full of eggs, you could make only two batches of brownies.
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Figure 3.12 The Concept of a Limiting Reactant in the Preparation of Brownies
Let’s now turn to a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride (TiCl 4) and carbon dioxide. Titanium tetrachloride is then converted to metallic titanium by reaction with magnesium metal at high temperature: Equation 3.22 TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l) Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about $100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium.
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Suppose you have 1.00 kg of titanium tetrachloride and 200 g of magnesium metal. How much titanium metal can you produce according to Equation 3.22? Solving this type of problem requires that you carry out the following steps: 1. Determine the number of moles of each reactant. 2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting. 3. Calculate the number of moles of product that can be obtained from the limiting reactant. 4. Convert the number of moles of product to mass of product.
1. To determine the number of moles of reactants present, you must calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows:
moles TiCl4 =
mass TiCl4 molar mass TiCl4
= 1000 g TiCl4 × moles Mg
=
mass Mg molar mass Mg
= 200 g Mg ×
Medical use of titanium. Here is an example of its successful use in joint replacement implants.
1 mol TiCl4 189.679 g TiCl4
1 mol Mg
24.305 g Mg
= 5.272 mol TiCl4
= 8.23 mol Mg
2. You have more moles of magnesium than of titanium tetrachloride, but the ratio is only
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mol Mg 8.23 mol = = 1.56 mol TiCl4 5.272 mol Because the ratio of the coefficients in the balanced chemical equation is
2 mol Mg = 2 1 mol TiCl4 you do not have enough magnesium to react with all the titanium tetrachloride. If this point is not clear from the mole ratio, you should calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, you have 8.23 mol of Mg, so you need (8.23 ÷ 2) = 4.12 mol of TiCl 4 for complete reaction. Because you have 5.272 mol of TiCl4, titanium tetrachloride is present in excess. Conversely, 5.272 mol of TiCl4 requires 2 × 5.272 = 10.54 mol of Mg, but you have only 8.23 mol. So magnesium is the limiting reactant. 3. Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed:
moles Ti = 8.23 mol Mg =
1 mol Ti 2 mol Mg
= 4.12 mol Ti
Thus only 4.12 mol of Ti can be formed. 4. To calculate the mass of titanium metal that you can obtain, multiply the number of moles of titanium by the molar mass of titanium (47.867 g/mol):
moles Ti = mass Ti × molar mass Ti = 4.12 mol Ti ×
47.867 g Ti 1 mol Ti
Here is a simple and reliable way to identify the limiting reactant in any problem of this sort: 1. Calculate the number of moles of each reactant present: 5.272 mol of TiCl4 and 8.23 mol of Mg.
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2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation:
TiCl4 :
5.272 mol (actual) 8.23 mol (actual) = 5.272 Mg: = 4.12 1 mol (stoich) 2 mol (stoich)
3. The reactant with the smallest mole ratio is limiting. Magnesium, with a calculated stoichiometric mole ratio of 4.12, is the limiting reactant. As you learned in Chapter 1 "Introduction to Chemistry", density is the mass per unit volume of a substance. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example 12 demonstrates.
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EXAMPLE 12 Ethyl acetate (CH3CO2C2H5) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol (C2H5OH) with acetic acid (CH3CO2H); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively.
Given: reactants, products, and volumes and densities of reactants Asked for: mass of product Strategy: A Balance the chemical equation for the reaction. B Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles. C Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product. D Convert from moles of product to mass of product. Solution:
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A We always begin by writing the balanced chemical equation for the reaction: C2H5OH(l) + CH3CO2H(aq) → CH3CO2C2H5(aq) + H2O(l) B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall from Chapter 1 "Introduction to Chemistry" that the density of a substance is the mass divided by the volume:
density =
mass volume
Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm 3):
moles C2 H5 OH
mass C2 H5 OH molar mass C2 H5 OH volume C 2 H5 OH × density C 2 H5 OH = molar mass C2 H5 OH =
= 10.0 mL C2 H5 OH ×
moles CH3 CO2 H
0.7893 g C2 H5 OH 1 mL C2 H5 OH
1
46.
= 0.171 mol C 2 H5 OH mass CH3 CO2 H = molar mass CH3 CO2 H volume CH 3 CO2 H × density CH 3 CO2 H = molar mass CH3 CO2 H = 10.0 mL CH3 CO2 H ×
1.0492 g CH3 CO2 H 1 mL CH3 CO2 H
= 0.175 mol CH 3 CO2 H C The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and
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0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol:
moles ethyl acetate
1 mol ethyl acetate 1 mol ethanol 1 mol CH 3 CO2 C2 H5 = 0.171 mol C2 H5 OH × 1 mol C2 H5 OH = mol ethanol ×
= 0.171 mol CH 3 CO2 C2 H5 D The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass:
mass of ethyl acetate
= mol ethyl acetate × molar mass ethyl acetate 88.11 g CH 3 CO = 0.171 mol CH3 CO2 C2 H5 × 1 mol CH3 CO2 = 15.1 g CH 3 CO2 C2 H5
Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm3) to determine the volume of ethyl acetate that could be produced:
volume of ethyl acetate
= 15.1 g CH3 CO2 C2 H5 ×
1 mL CH3 CO2
0.9003 g CH3 CO
= 16.8 mL CH3 CO2 C2 H5 Exercise Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound P4S10. How much P4S10 can be prepared starting with 10.0 g of P4 and 30.0 g of S8? Answer: 35.9 g
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Percent Yield You have learned that when reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yield18, the amount you would obtain if the reaction occurred perfectly and your method of purifying the product were 100% efficient. In reality, you almost always obtain less product than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not go all the way to completion, thus resulting in a mixture of products and reactants. This last possibility is a common occurrence and is the subject of Chapter 15 "Chemical Equilibrium". So the actual yield19, the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yield20 of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage: Equation 3.23
percent yield =
actual yield (g) × 100 theoretical yield (g)
The method used to calculate the percent yield of a reaction is illustrated in Example 13. 18. The maximum amount of product that can be formed from the reactants in a chemical reaction, which theoretically is the amount of product that would be obtained if the reaction occurred perfectly and the method of purifying the product were 100% efficient. 19. The measured mass of products actually obtained from a reaction. The actual yield is nearly always less than the theoretical yield. 20. The ratio of the actual yield of a reaction to the theoretical yield multiplied by 100 to give a percentage.
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EXAMPLE 13 Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction
C7 H7 NO2
p-aminobenzoic acid
+
C6 H15 NO
2-diethylaminoethanol
H SO
2 ⎯⎯⎯⎯⎯ →4 C13 H20 N2 O2 + H2 O
procaine
If we carried out this reaction using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and we isolated 15.7 g of procaine, what was the percent yield?
The preparation of procaine. A reaction of p-aminobenzoic acid with 2-diethylaminoethanol yields procaine and water.
Given: masses of reactants and product Asked for: percent yield Strategy: A Write the balanced chemical equation. B Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based
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on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield. C Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100. Solution: A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water. B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/ mol. Thus the reaction used the following numbers of moles of reactants:
moles p-aminobenzoic acid = 10.0 g ×
1 mol = 0.0729 mol p-am 137.14 g
moles 2-diethylaminoethanol = 10.0 g ×
1 mol = 0.0853 mol 2117.19 g
The reaction requires a 1:1 mole ratio of the two reactants, so paminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol.
theoretical yield of procaine = 0.0729 mol ×
236.31 g 1 mol
= 17.2 g
C The actual yield was only 15.7 g of procaine, so the percent yield was
percent yield =
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15.7 g × 100 = 91.3% 17.2 g
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(If the product were pure and dry, this yield would indicate that we have very good lab technique!) Exercise Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction: 2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g) The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows: PbO(s) + C(s) → Pb(l) + CO(g) If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?
Crystalline galena (a) and a sample of lead (b). Pure lead is soft enough to be shaped easily with a hammer, unlike the brittle mineral galena, the main ore of lead.
Answer: 89.2%
Percent yield can range from 0% to 100%.In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually
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happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments! A 100% yield means that everything worked perfectly, and you obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows how unlikely a 100% yield is. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.
Summary The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
KEY TAKEAWAY • The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained.
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CONCEPTUAL PROBLEMS 1. Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer. 2. Given the equation 2H2(g) + O2(g) → 2H2O(g), is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor? 3. What does it mean to say that a reaction is stoichiometric? 4. When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer. 5. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems. 1. What is the formula mass of each species? a. b. c. d. e. f.
ammonium chloride sodium cyanide magnesium hydroxide calcium phosphate lithium carbonate hydrogen sulfite ion
2. What is the molecular or formula mass of each compound? a. b. c. d. e. f.
potassium permanganate sodium sulfate hydrogen cyanide potassium thiocyanate ammonium oxalate lithium acetate
3. How many moles are in each of the following? a. b. c. d. e. f.
10.76 g of Si 8.6 g of Pb 2.49 g of Mg 0.94 g of La 2.68 g of chlorine gas 0.089 g of As
4. How many moles are in each of the following? a. b. c. d. e. f.
8.6 g of CO2 2.7 g of CaO 0.89 g of KCl 4.3 g of SrBr2 2.5 g of NaOH 1.87 g of Ca(OH)2
5. Convert the following to moles and millimoles. a. 1.68 g of Ba(OH)2 b. 0.792 g of H3PO4 c. 3.21 g of K2S
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d. e. f. g. h.
0.8692 g of Cu(NO3)2 10.648 g of Ba3(PO4)2 5.79 g of (NH4)2SO4 1.32 g of Pb(C2H3O2)2 4.29 g of CaCl2·6H2O
6. Convert the following to moles and millimoles. a. b. c. d. e. f. g. h. i.
0.089 g of silver nitrate 1.62 g of aluminum chloride 2.37 g of calcium carbonate 1.004 g of iron(II) sulfide 2.12 g of dinitrogen pentoxide 2.68 g of lead(II) nitrate 3.02 g of ammonium phosphate 5.852 g of sulfuric acid 4.735 g of potassium dichromate
7. What is the mass of each substance in grams and milligrams? a. b. c. d. e. f.
5.68 mol of Ag 2.49 mol of Sn 0.0873 mol of Os 1.74 mol of Si 0.379 mol of H2 1.009 mol of Zr
8. What is the mass of each substance in grams and milligrams? a. b. c. d. e. f.
2.080 mol of CH3OH 0.288 mol of P4 3.89 mol of ZnCl2 1.800 mol of Fe(CO)5 0.798 mol of S8 4.01 mol of NaOH
9. What is the mass of each compound in kilograms? a. b. c. d. e. f.
6.38 mol of P4O10 2.26 mol of Ba(OH)2 4.35 mol of K3PO4 2.03 mol of Ni(ClO3)2 1.47 mol of NH4NO3 0.445 mol of Co(NO3)3
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10. How many atoms are contained in each? a. b. c. d. e. f. g.
2.32 mol of Bi 0.066 mol of V 0.267 mol of Ru 4.87 mol of C 2.74 g of I2 1.96 g of Cs 7.78 g of O2
11. Convert each number of atoms to milligrams. a. 5.89 × 1022 Pt atoms b. 2.899 × 1021 Hg atoms c. 4.826 × 1022 atoms of chlorine 12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess. a. 2.46 g barium(s) plus 3.89 g bromine(l) in water to give barium bromide b. 1.44 g bromine(l) plus 2.42 g potassium iodide(s) in water to give potassium bromide and iodine c. 1.852 g of Zn metal plus 3.62 g of sulfuric acid in water to give zinc sulfate and hydrogen gas d. 0.147 g of iron metal reacts with 0.924 g of silver acetate in water to give iron(II) acetate and silver metal e. 3.142 g of ammonium phosphate reacts with 1.648 g of barium hydroxide in water to give ammonium hydroxide and barium phosphate 13. Under the proper conditions, ammonia and oxygen will react to form dinitrogen monoxide (nitrous oxide, also called laughing gas) and water. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants. a. b. c. d.
24.6 g of ammonia and 21.4 g of oxygen 3.8 mol of ammonia and 84.2 g of oxygen 3.6 × 1024 molecules of ammonia and 318 g of oxygen 2.1 mol of ammonia and 36.4 g of oxygen
14. When a piece of zinc metal is placed in aqueous hydrochloric acid, zinc chloride is produced, and hydrogen gas is evolved. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants. a. 12.5 g of HCl and 7.3 g of Zn b. 6.2 mol of HCl and 100 g of Zn
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c. 2.1 × 1023 molecules of Zn and 26.0 g of HCl d. 3.1 mol of Zn and 97.4 g of HCl 15. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced. a. NaI(aq) + Cl2(g) → NaCl(aq) + I2(s); 1.0 mol of NaCl b. NaCl(aq) + H2SO4(aq) → HCl(g) + Na2SO4(aq); 0.50 mol of HCl c. NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq); 1.5 mol of HNO3 16. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced. a. AgNO3(aq) + CaCl2(s) → AgCl(s) + Ca(NO3)2(aq); 1.25 mol of AgCl b. Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s) + H2O(l); 3.8 g of PbSO4 c. H3PO4(aq) + MgCO3(s) → Mg3(PO4)2(s) + CO2(g) + H2O(l); 6.41 g of Mg3(PO4)2 17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.) a.
KClO 3 (s) ⎯→ KCl (s) + O 2 (g );2.14 g of KClO3 produces 0.67 g of Δ
O2 b. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide c. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate 18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction? 19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction? 20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting. A + 3B → 2C 21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol. a. Write a balanced chemical equation for this reaction. b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide.
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22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction: NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq) a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2? b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide. 23. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield. a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion? b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane? c. Use the actual percent yield to calculate how many grams of each product would actually be obtained. 24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions: Δ
2PaI5 (s) ⎯→ 2Pa(s) + 5I 2 (s) a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized? b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction? c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation? 25. Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction: C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s) Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia. a. b. c. d. e.
Which is the limiting reactant? Which reactant is present in excess? What is the theoretical yield of ammonium chloride in grams? If 4.78 g of NH4Cl was recovered, what was the percent yield? Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess.
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26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following: a. 9.36 × 1024 formula units of NaCl b. 8.5 × 104 mol of Br2 c. 3.7 × 108 g of NaCl
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ANSWERS 1. a. b. c. d. e. f.
53.941 amu 49.0072 amu 58.3197 amu 310.177 amu 73.891 amu 81.071 amu
3. a. b. c. d. e. f.
0.3831 mol Si 4.2 × 10−2 mol Pb 0.102 mol Mg 6.8 × 10−3 mol La 3.78 × 10−2 mol Cl2 1.2 × 10−3 mol As
5. a. b. c. d. e. f. g. h.
9.80 × 10−3 mol or 9.80 mmole Ba(OH)2 8.08 × 10−3 mol or 8.08 mmole H3PO4 2.91 × 10−2 mol or 29.1 mmole K2S 4.634 × 10−3 mol or 4.634 mmole Cu(NO3)2 1.769 × 10−2 mol 17.69 mmole Ba3(PO4)2 4.38 × 10−2 mol or 43.8 mmole (NH4)2SO4 4.06 × 10−3 mol or 4.06 mmole Pb(C2H3O2)2 1.96 × 10−2 mol or 19.6 mmole CaCl2· 6H2O
7. a. b. c. d. e. f.
613 g or 6.13 × 105 mg Ag 296 g or 2.96 × 105 mg Sn 16.6 g or 1.66 × 104 mg Os 48.9 g or 4.89 × 104 mg Si 0.764 g or 764 mg H2 92.05 g or 9.205 × 104 mg Zr
9. a. b. c. d. e. f. 11.
1.81 kg P4O10 0.387 kg Ba(OH)2 0.923 kg K3PO4 0.458 kg Ni(ClO3)2 0.118 kg (NH4)NO3 0.109 kg Co(NO3)3
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a. 1.91 × 104 mg Pt b. 965.6 mg Hg c. 2841 mg Cl 13. The balanced chemical equation for this reaction is 2NH3 + 2O2 → N2O + 3H2O a. b. c. d.
NH3 NH3 O2 NH3
15. a. 150 g NaI and 35 g Cl2 b. 29 g NaCl and 25 g H2SO4 c. 140 g NO2 and 27 g H2O 17. a. 80% b. 30% c. 35.7% 19. 45%. 21. a. CO + 2H2 → CH3OH b. 58.28% 23. a. 2.24 g Cl2 b. 4.95 g c. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3 25. a. b. c. d.
chlorobenzene ammonia 8.74 g ammonium chloride. 55%
e. Theoretical yield (NH4Cl) =
mass of chlorobenzene (g) × 0.92 × 53.49 g/mol 112.55 g/mol
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3.5 Classifying Chemical Reactions LEARNING OBJECTIVES 1. To identify fundamental types of chemical reactions. 2. To predict the types of reactions substances will undergo.
The chemical reactions we have described are only a tiny sampling of the infinite number of chemical reactions possible. How do chemists cope with this overwhelming diversity? How do they predict which compounds will react with one another and what products will be formed? The key to success is to find useful ways to categorize reactions. Familiarity with a few basic types of reactions will help you to predict the products that form when certain kinds of compounds or elements come in contact.
21. A reaction of the general form acid + base → salt. 22. A chemical reaction that has the general form AB + C → AC + B or AB + CD → AD + CB. 23. A chemical reaction that has the general form A + B → AB. Condensation reactions are the reverse of cleavage reactions. Some, but not all, condensation reactions are also oxidation–reduction reactions.
Most chemical reactions can be classified into one or more of five basic types: acid–base reactions21, exchange reactions22, condensation reactions23 (and the reverse, cleavage reactions24), and oxidation–reduction reactions25. The general forms of these five kinds of reactions are summarized in Table 3.1 "Basic Types of Chemical Reactions", along with examples of each. It is important to note, however, that many reactions can be assigned to more than one classification, as you will see in our discussion. The classification scheme is only for convenience; the same reaction can be classified in different ways, depending on which of its characteristics is most important. Oxidation–reduction reactions, in which there is a net transfer of electrons from one atom to another, and condensation reactions are discussed in this section. Acid–base reactions and one kind of exchange reaction—the formation of an insoluble salt such as barium sulfate when solutions of two soluble salts are mixed together—will be discussed in Chapter 4 "Reactions in Aqueous Solution".
24. A chemical reaction that has the general form AB → A + B. Cleavage reactions are the reverse of condensation reactions. 25. A chemical reaction that exhibits a change in the oxidation states of one or more elements in the reactants that has the general form oxidant + reductant → reduced oxidant + oxidized reductant.
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Table 3.1 Basic Types of Chemical Reactions Name of Reaction
General Form
oxidation–reduction oxidant + reductant → reduced (redox) oxidant + oxidized reductant acid–base
exchange
Example(s) C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
acid + base → salt
NH3(aq) + HNO3(aq) → NH4+(aq) + NO3−(aq)
AB + C → AC + B
CH3Cl + OH− → CH3OH + Cl−
AB + CD → AD + CB
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) CO2(g) + H2O(l) → H2CO3(aq)
condensation
A + B → AB
HBr + H2C=CH2 → CH3CH2Br* CaCO3(s) → CaO(s) + CO2(g)
cleavage
AB → A + B
CH3CH2Cl → H2C=CH2 + HCl**
*
In more advanced chemisty courses you will learn that this reaction is also called an addition reaction.
**
In more advanced chemistry courses you will learn that this reaction is also called an elimination reaction.
Oxidation–Reduction Reactions The term oxidation26 was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O 2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. 26. The loss of one or more electrons in a chemical reaction. The substance that loses electrons is said to be oxidized.
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Note the Pattern Any oxidation must be accompanied by a reduction and vice versa.
Originally, the term reduction27 referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). The reaction is as follows: Equation 3.24 Cu2O(s) + H2(g) → 2Cu(s) + H2O(g) Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state28 of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is Equation 3.25 4Al(s) + 3O2(g) → 2Al2O3(s)
27. The gain of one or more electrons in a chemical reaction. The substance that gains electrons is said to be reduced. 28. The charge that each atom in a compound would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons.
Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the superscript 0 emphasizes the oxidation state of the elements): Equation 3.26 4Al0 + 3O20 → 4Al3+ + 6O2−
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Equation 3.24 and Equation 3.25 are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation 3.26, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen: Equation 3.27
electrons lost = 4 Al atoms ×
3 e− lost
= 12 e − lost
Al atom 2 e− gained electrons gained = 6 O atoms × = 12 e − gained O atom The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in Figure 3.1 "Samples of 1 Mol of Some Common Substances".
Note the Pattern In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.
Assigning Oxidation States Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. In Chapter 2 "Molecules, Ions, and Chemical Formulas", you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure 3.13 "The Reaction of a Neutral Sodium Atom with a Neutral Chlorine Atom"), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share
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electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions. A set of rules for assigning oxidation states to atoms in chemical compounds follows. As we discuss atomic and molecular structure in Chapter 6 "The Structure of Atoms", Chapter 7 "The Periodic Table and Periodic Trends", Chapter 8 "Ionic versus Covalent Bonding", and Chapter 9 "Molecular Geometry and Covalent Bonding Models", the principles underlying these rules will be described more fully.
Figure 3.13 The Reaction of a Neutral Sodium Atom with a Neutral Chlorine Atom
Rules for Assigning Oxidation StatesNonintegral oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states. 1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero. 2. The oxidation state of a monatomic ion is The result is the transfer of one + the same as its charge—for example, Na = electron from sodium to chlorine, forming the ionic compound +1, Cl− = −1. NaCl. 3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens. 4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals. 5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present. 6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion. In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule
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or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound. Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6 "The Structure of Atoms". Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H− ion, whereas HCl forms H+ and Cl− ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable. The reduction of copper(I) oxide shown in Equation 3.28 demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows: Equation 3.28 +1
0
0
+1
Cu 2 O(s) + H2 (g) → 2Cu (s) + H2 O(g) –2
–2
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge: Equation 3.29
electrons lost = 2 H atoms × electrons gained = 2 Cu atoms ×
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1 e− lost
= 2 e− lost
H atom 1 e− gained Cu atom
= 2 e− gained
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Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.
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EXAMPLE 14 Assign oxidation states to all atoms in each compound. a. b. c. d. e.
sulfur hexafluoride (SF6) methanol (CH3OH) ammonium sulfate [(NH4)2SO4] magnetite (Fe3O4) ethanoic (acetic) acid (CH3CO2H)
Given: molecular or empirical formula Asked for: oxidation states Strategy: Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1. Solution:
a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6: [(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0 b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2: [(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
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c. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42−) (see Table 2.4 "Common Polyatomic Ions and Their Names"). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3: [(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6: [(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:
8 [(4 O atoms)( − 2)] + [(3 Fe atoms) (+ 3 )] = 0 Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.” e. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of [(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0
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So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO 2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of [(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3 To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus +1 +3 +1 C H3 C O2 H −3 −2
Thus the sum of the oxidation states of the two carbon atoms is indeed zero. Exercise Assign oxidation states to all atoms in each compound. a. b. c. d.
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barium fluoride (BaF2) formaldehyde (CH2O) potassium dichromate (K2Cr2O7) cesium oxide (CsO2)
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e. ethanol (CH3CH2OH) Answer: a. b. c. d. e.
Ba, +2; F, −1 C, 0; H, +1; O, −2 K, +1; Cr, +6; O, −2 Cs, +1; O, −½ C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Oxidants and Reductants Compounds that are capable of accepting electrons, such as O 2 or F2, are called oxidants (or oxidizing agents)29 because they can oxidize other compounds. In the process of accepting electrons, an oxidant is reduced. Compounds that are capable of donating electrons, such as sodium metal or cyclohexane (C 6H12), are called reductants (or reducing agents)30 because they can cause the reduction of another compound. In the process of donating electrons, a reductant is oxidized. These relationships are summarized in Equation 3.30: Equation 3.30 oxidant + reductant → oxidation−reduction
O2 (g) + 4Na(s) → 2Na 2 O(s)
gains e –
loses e –
(is reduced)
(is oxidized)
redox reaction
30. A compound that is capable of donating electrons; thus it is oxidized.
Some oxidants have a greater ability than others to remove electrons from other compounds. Oxidants can range from very powerful, capable of oxidizing most compounds with which they come in contact, to rather weak. Both F 2 and Cl2 are powerful oxidants: for example, F2 will oxidize H2O in a vigorous, potentially explosive reaction. In contrast, S8 is a rather weak oxidant, and O2 falls somewhere in between. Conversely, reductants vary in their tendency to donate electrons to other compounds. Reductants can also range from very powerful, capable of giving up electrons to almost anything, to weak. The alkali metals are powerful reductants, so they must be kept away from atmospheric oxygen to avoid a potentially hazardous redox reaction.
31. An oxidation–reduction reaction in which the oxidant is O2 .
A combustion reaction31, first introduced in Section 3.2 "Determining Empirical and Molecular Formulas", is an oxidation–reduction reaction in which the oxidant
29. A compound that is capable of accepting electrons; thus it is reduced.
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is O2. One example of a combustion reaction is the burning of a candle, shown in Figure 3.9 "An Example of a Combustion Reaction". Consider, for example, the combustion of cyclohexane, a typical hydrocarbon, in excess oxygen. The balanced chemical equation for the reaction, with the oxidation state shown for each atom, is as follows: Equation 3.31 +1
0
+4
+1
C6 H12 + 9O2 → 6C O2 + 6H2 O –2
–2
–2
If we compare the oxidation state of each element in the products and the reactants, we see that hydrogen is the only element whose oxidation state does not change; it remains +1. Carbon, however, has an oxidation state of −2 in cyclohexane and +4 in CO2; that is, each carbon atom changes its oxidation state by six electrons during the reaction. Oxygen has an oxidation state of 0 in the reactants, but it gains electrons to have an oxidation state of −2 in CO2 and H2O. Because carbon has been oxidized, cyclohexane is the reductant; because oxygen has been reduced, it is the oxidant. All combustion reactions are therefore oxidation–reduction reactions.
Condensation Reactions The reaction of bromine with ethylene to give 1,2-dibromoethane, which is used in agriculture to kill nematodes in soil, is as follows: Equation 3.32 C2H4(g) + Br2(g) → BrCH2CH2Br(g) According to Table 3.1 "Basic Types of Chemical Reactions", this is a condensation reaction because it has the general form A + B → AB. This reaction, however, can also be viewed as an oxidation–reduction reaction, in which electrons are transferred from carbon (−2 → −1) to bromine (0 → −1). Another example of a condensation reaction is the one used for the industrial synthesis of ammonia: Equation 3.33 3H2(g) + N2(g) → 2NH3(g)
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Although this reaction also has the general form of a condensation reaction, hydrogen has been oxidized (0 → +1) and nitrogen has been reduced (0 → −3), so it can also be classified as an oxidation–reduction reaction. Not all condensation reactions are redox reactions. The reaction of an amine with a carboxylic acid, for example, is a variant of a condensation reaction (A + B → A′B′ + C): two large fragments condense to form a single molecule, and a much smaller molecule, such as H2O, is eliminated. In this reaction, the −OH from the carboxylic acid group and −H from the amine group are eliminated as H 2O, and the reaction forms an amide bond (also called a peptide bond) that links the two fragments. Amide bonds are the essential structural unit linking the building blocks of proteins and many polymers together, as described in Chapter 12 "Solids". Nylon, for example, is produced from a condensation reaction (Figure 3.14 "The Production of Nylon").
Amide bonds. The reaction of an amine with a carboxylic acid proceeds by eliminating water and forms a new C–N (amide) bond.
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Figure 3.14 The Production of Nylon
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EXAMPLE 15 The following reactions have important industrial applications. Using Table 3.1 "Basic Types of Chemical Reactions", classify each reaction as an oxidation–reduction reaction, an acid–base reaction, an exchange reaction, a condensation reaction, or a cleavage reaction. For each redox reaction, identify the oxidant and reductant and specify which atoms are oxidized or reduced. (Don’t forget that some reactions can be placed into more than one category.) a. b. c. d. e.
C2H4(g) + Cl2(g) → ClCH2CH2Cl(g) AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) CaCO3(s) → CaO(s) + CO2(g) Ca5(PO4)3(OH)(s) + 7H3PO4(aq) + 4H2O(l) → 5Ca(H2PO4)2·H2O(s) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
Given: balanced chemical equation Asked for: classification of chemical reaction Strategy: A Determine the general form of the equation by referring to Table 3.1 "Basic Types of Chemical Reactions" and then classify the reaction. B For redox reactions, assign oxidation states to each atom present in the reactants and the products. If the oxidation state of one or more atoms changes, then the reaction is a redox reaction. If not, the reaction must be one of the other types of reaction listed in Table 3.1 "Basic Types of Chemical Reactions". Solution:
a. A This reaction is used to prepare 1,2-dichloroethane, one of the top 25 industrial chemicals in Figure 2.22 "Top 25 Chemicals Produced in the United States in 2002*". It has the general form A + B → AB, which is typical of a condensation reaction. B Because reactions may fit into more than one category, we need to look at the oxidation states of the atoms:
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+1
0
+1
+1
C 2 H 4 + Cl2 → Cl C H 2 C H 2 Cl –1 –1
–2
–1
–1
The oxidation states show that chlorine is reduced from 0 to −1 and carbon is oxidized from −2 to −1, so this is a redox reaction as well as a condensation reaction. Ethylene is the reductant, and chlorine is the oxidant. b. A This reaction is used to prepare silver chloride for making photographic film. The chemical equation has the general form AB + CD → AD + CB, so it is classified as an exchange reaction. B The oxidation states of the atoms are as follows +1 +5
+1
+1
+1 +5
Ag N O 3 + Na Cl → Ag Cl + Na N O 3 –2
–1
–1
–2
There is no change in the oxidation states, so this is not a redox reaction.
AgCl(s) precipitates when solutions of AgNO3(aq) and NaCl(aq) are mixed. NaNO3 (aq) is in solution as Na+ and NO3− ions.
c. A This reaction is used to prepare lime (CaO) from limestone (CaCO3) and has the general form AB → A + B. The chemical equation’s general form indicates that it can be classified as a cleavage reaction, the reverse of a condensation reaction. B The oxidation states of the atoms are as follows:
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+2 +4
+2
+4
Ca C O 3 → CaO + C O 2 –2
–2
–2
Because the oxidation states of all the atoms are the same in the products and the reactant, this is not a redox reaction. d. A This reaction is used to prepare “super triple phosphate” in fertilizer. One of the reactants is phosphoric acid, which transfers a proton (H+) to the phosphate and hydroxide ions of hydroxyapatite [Ca5(PO4)3(OH)] to form H2PO4− and H2O, respectively. This is an acid–base reaction, in which H 3PO4 is the acid (H+ donor) and Ca5(PO4)3(OH) is the base (H+ acceptor). B To determine whether it is also a redox reaction, we assign oxidation states to the atoms: +2 +5 +1 Ca5 (PO 4 )3 (O H ) –2 –2
+
+1 +5 7H 3 P O 4 –2
+1
+ 4H 2 O → –2
+2 +1 +5 5Ca(H 2 P O 4 ) –2 2
·
+1 H2 O –2
Because there is no change in oxidation state, this is not a redox reaction. e. A This reaction occurs in a conventional car battery every time the engine is started. An acid (H2SO4) is present and transfers protons to oxygen in PbO2 to form water during the reaction. The reaction can therefore be described as an acid–base reaction. B The oxidation states are as follows: 0
+4
+2 +6
+1 +6
+1
Pb + Pb O 2 + 2H 2 S O 4 → 2Pb S O 4 + 2H 2 O –2
–2
–2
–2
The oxidation state of lead changes from 0 in Pb and +4 in PbO 2 (both reactants) to +2 in PbSO4. This is also a redox reaction, in which elemental lead is the reductant, and PbO2 is the oxidant. Which description is correct? Both.
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Schematic drawing of a 12-volt car battery. The locations of the reactants (lead metal in a spongy form with large surface area) and PbO2 are shown. The product (PbSO4) forms as a white solid between the plates.
Exercise Using Table 3.1 "Basic Types of Chemical Reactions", classify each reaction as an oxidation–reduction reaction, an acid–base reaction, an exchange reaction, a condensation reaction, or a cleavage reaction. For each redox reaction, identify the oxidant and the reductant and specify which atoms are oxidized or reduced. a. b. c. d. e.
Al(s) + OH−(aq) + 3H2O(l) → 3/2H2(g) + [Al(OH)4]−(aq) TiCl4(l) + 2Mg(l) → Ti(s) + 2MgCl2(l) MgCl2(aq) + Na2CO3(aq) → MgCO3(s) + 2NaCl(aq) CO(g) + Cl2(g) → Cl2CO(l) H2SO4(l) + 2NH3(g) → (NH4)2SO4(s)
Answer: a. Redox reaction; reductant is Al, oxidant is H2O; Al is oxidized, H is reduced. This is the reaction that occurs when Drano is used to clear a clogged drain.
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b. Redox reaction; reductant is Mg, oxidant is TiCl4; Mg is oxidized, Ti is reduced. c. Exchange reaction. This reaction is responsible for the scale that develops in coffee makers in areas that have hard water. d. Both a condensation reaction and a redox reaction; reductant is CO, oxidant is Cl2; C is oxidized, Cl is reduced. The product of this reaction is phosgene, a highly toxic gas used as a chemical weapon in World War I. Phosgene is now used to prepare polyurethanes, which are used in foams for bedding and furniture and in a variety of coatings. e. Acid–base reaction.
Catalysts Many chemical reactions, including some of those discussed previously, occur more rapidly in the presence of a catalyst32, which is a substance that participates in a reaction and causes it to occur more rapidly but can be recovered unchanged at the end of a reaction and reused. Because catalysts are not involved in the stoichiometry of a reaction, they are usually shown above the arrow in a net chemical equation. Chemical processes in industry rely heavily on the use of catalysts, which are usually added to a reaction mixture in trace amounts, and most biological reactions do not take place without a biological catalyst or enzyme33. Examples of catalyzed reactions in industry are the use of platinum in petroleum cracking and reforming, the reaction of SO2 and O2 in the presence of V2O5 to produce SO3 in the industrial synthesis of sulfuric acid, and the use of sulfuric acid in the synthesis of compounds such as ethyl acetate and procaine. Not only do catalysts greatly increase the rates of reactions, but in some cases such as in petroleum refining, they also control which products are formed. The acceleration of a reaction by a catalyst is called catalysis34.
32. A substance that increases the rate of a chemical reaction without undergoing a net chemical change itself. 33. Catalysts that occur naturally in living organisms and catalyze biological reactions. 34. The acceleration of a chemical reaction by a catalyst.
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Catalysts may be classified as either homogeneous or heterogeneous. A homogeneous catalyst35 is uniformly dispersed throughout the reactant mixture to form a solution. Sulfuric acid, for example, is a homogeneous catalyst used in the synthesis of esters such as procaine (Example 13). An ester has a structure similar to that of a carboxylic acid, in which the hydrogen atom attached A heterogeneous catalyst. This to oxygen has been replaced by an R group. They are large circular gauze, woven from responsible for the fragrances of many fruits, flowers, rhodium-platinum wire, is a heterogeneous catalyst in the and perfumes. Other examples of homogeneous commercial production of nitric catalysts are the enzymes that allow our bodies to acid by the oxidation of 36 function. In contrast, a heterogeneous catalyst is in a ammonia. different physical state than the reactants. For economic reasons, most industrial processes use heterogeneous catalysts in the form of solids that are added to solutions of the reactants. Because such catalysts often contain expensive precious metals such as platinum or palladium, it makes sense to formulate them as solids that can be easily separated from the liquid or gaseous reactant-product mixture and recovered. Examples of heterogeneous catalysts are the iron oxides used in the industrial synthesis of ammonia and the catalytic converters found in virtually all modern automobiles, which contain precious metals like palladium and rhodium. Catalysis will be discussed in more detail in Chapter 14 "Chemical Kinetics" when we discuss reaction rates, but you will encounter the term frequently throughout the text.
35. A catalyst that is uniformly dispersed throughout the reactant mixture to form a solution. 36. A catalyst that is in a different physical state than the reactants.
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Summary Chemical reactions may be classified as an acid–base reaction, an exchange reaction, a condensation reaction and its reverse, a cleavage reaction, and an oxidation–reduction (or redox) reaction. To keep track of electrons in chemical reactions, oxidation states are assigned to atoms in compounds. The oxidation state is the charge an atom would have if all its bonding electrons were transferred completely to the atom that has the greater attraction for electrons. In an oxidation–reduction reaction, one atom must lose electrons and another must gain electrons. Oxidation is the loss of electrons, and an element whose oxidation state increases is said to be oxidized. Reduction is the gain of electrons, and an element whose oxidation state decreases is said to be reduced. Oxidants are compounds that are capable of accepting electrons from other compounds, so they are reduced during an oxidation–reduction reaction. In contrast, reductants are compounds that are capable of donating electrons to other compounds, so they are oxidized during an oxidation–reduction reaction. A combustion reaction is a redox reaction in which the oxidant is O2(g). An amide bond is formed from the condensation reaction between a carboxylic acid and an amine; it is the essential structural unit of proteins and many polymers. A catalyst is a substance that increases the rate of a chemical reaction without undergoing a net chemical change itself. A biological catalyst is called an enzyme. Catalysis is an acceleration in the rate of a reaction caused by the presence of a substance that does not appear in the chemical equation. A homogeneous catalyst is uniformly dispersed in a solution of the reactants, whereas a heterogeneous catalyst is present as a different phase, usually a solid.
KEY TAKEAWAY • Chemical reactions may be classified as acid–base, exchange, condensation, cleavage, and oxidation–reduction (redox).
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CONCEPTUAL PROBLEMS 1. What is a combustion reaction? How can it be distinguished from an exchange reaction? 2. What two products are formed in the combustion of an organic compound containing only carbon, hydrogen, and oxygen? Is it possible to form only these two products from a reaction that is not a combustion reaction? Explain your answer. 3. What factors determine whether a reaction can be classified as a redox reaction? 4. Name three characteristics of a balanced redox reaction. 5. Does an oxidant accept electrons or donate them? 6. Does the oxidation state of a reductant become more positive or more negative during a redox reaction? 7. Nitrogen, hydrogen, and ammonia are known to have existed on primordial earth, yet mixtures of nitrogen and hydrogen do not usually react to give ammonia. What natural phenomenon would have enough energy to initiate a reaction between these two primordial gases? 8. Catalysts are not added to reactions in stoichiometric quantities. Why? 9. State whether each of the following uses a homogeneous catalyst or a heterogeneous catalyst. a. Platinum metal is used in the catalytic converter of an automobile. b. Nitrogen is biologically converted to ammonia by an enzyme. c. Carbon monoxide and hydrogen combine to form methane and water with a nickel catalyst. d. A dissolved rhodium compound is used as a catalyst for the conversion of an alkene to an alkane. 10. State whether each of the following uses a homogeneous catalyst or a heterogeneous catalyst. a. Pellets of ZSM-5, an aluminum- and silicon-containing mineral, are used to catalyze the conversion of methanol to gasoline. b. The conversion of glucose to a carboxylic acid occurs with catalysis by the enzyme glucose oxidase. c. Metallic rhodium is used to the conversion of carbon monoxide and water to carbon dioxide and hydrogen.
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11. Complete the following table to describe some key differences between homogeneous and heterogeneous catalysis. Homogeneous Heterogeneous number of phases ease of separation from product ease of recovery of catalyst 12. To increase the rate of a reaction, a scientist decided to use a catalyst. Unexpectedly, the scientist discovered that the catalyst decreased the yield of the desired product, rather than increasing it. What might have happened?
ANSWER Homogeneous Heterogeneous
11.
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number of phases
single phase
at least two phases
ease of separation from product
difficult
easy
ease of recovery of catalyst
difficult
easy
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems. 1. Classify each chemical reaction according to the types listed in Table 3.1 "Basic Types of Chemical Reactions". a. b. c. d.
12FeCl2(s) + 3O2(g) → 8FeCl3(s) + 2Fe2O3(s) CaCl2(aq) + K2SO4(aq) → CaSO4(s) + 2KCl(aq) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Br2(l) + C2H4(g) → BrCH2CH2Br(l)
2. Classify each chemical reaction according to the types listed in Table 3.1 "Basic Types of Chemical Reactions". a. b. c. d.
4FeO(s) + O2(g) → 2Fe2O3(s) Ca3(PO4)2(s) + 3H2SO4(aq) → 3CaSO4(s) + 2H3PO4(aq) HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l) ethane(g) + oxygen(g) → carbon dioxide(g) + water(g)
3. Assign oxidation states to the atoms in each compound or ion. a. b. c. d. e. f. g. h. i.
(NH4)2S the phosphate ion [AlF6]3− CuS HCO3− NH4+ H2SO4 formic acid n-butanol
4. Assign oxidation states to the atoms in each compound or ion. a. b. c. d. e. f. g. h. i.
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ClO2 HO2− sodium bicarbonate MnO2 PCl5 [Mg(H2O)6]2+ N 2O4 butanoic acid methanol
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5. Balance this chemical equation: NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l) What type of reaction is this? Justify your answer. 6. Assign oxidation states to the atoms in each compound. a. b. c. d. e. f. g. h. i.
iron(III) nitrate Al2O3 potassium sulfate Cr2O3 sodium perchlorate Cu2S hydrazine (N2H4) NO2 n-pentanol
7. Assign oxidation states to the atoms in each compound. a. b. c. d. e. f. g. h. i.
calcium carbonate NaCl CO2 potassium dichromate KMnO4 ferric oxide Cu(OH)2 Na2SO4 n-hexanol
8. For each redox reaction, determine the identities of the oxidant, the reductant, the species oxidized, and the species reduced. a. H2(g) + I2(s) → 2HI(g) b. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) c. 2F2(g) + 2NaOH(aq) → OF2(g) + 2NaF(aq) + H2O(l) 9. For each redox reaction, determine the identities of the oxidant, the reductant, the species oxidized, and the species reduced. a. 2Na(s) + Cl2(g) → 2NaCl(s) b. SiCl4(l) + 2Mg(s) → 2MgCl2(s) + Si(s) c. 2H2O2(aq) → 2H2O(l) + O2(g)
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10. Balance each chemical equation. Then identify the oxidant, the reductant, the species oxidized, and the species reduced. (Δ indicates that the reaction requires heating.) a. H2O(g) + CO(g) → CO2(g) + H2(g) b. the reaction of aluminum oxide, carbon, and chlorine gas at 900ºC to produce aluminum chloride and carbon monoxide c.
Δ
HgO(s) ⎯→ Hg(l) + O 2 (g)
11. Balance each chemical equation. Then identify the oxidant, the reductant, the species oxidized, and the species reduced. (Δ indicates that the reaction requires heating.) a. the reaction of water and carbon at 800ºC to produce hydrogen and carbon monoxide b. Mn(s) + S8(s) + CaO(s) → CaS(s) + MnO(s) c. the reaction of ethylene and oxygen at elevated temperature in the presence of a silver catalyst to produce ethylene oxide
d. ZnS(s) + H2SO4(aq) + O2(g) → ZnSO4(aq) + S8(s) + H2O(l) 12. Silver is tarnished by hydrogen sulfide, an atmospheric contaminant, to form a thin layer of dark silver sulfide (Ag2S) along with hydrogen gas. a. Write a balanced chemical equation for this reaction. b. Which species has been oxidized and which has been reduced? c. Assuming 2.2 g of Ag has been converted to silver sulfide, construct a table showing the reaction in terms of the number of atoms in the reactants and products, the moles of reactants and products, the grams of reactants and products, and the molecules of reactants and products. 13. The following reaction is used in the paper and pulp industry: Na2SO4(aq) + C(s) + NaOH(aq) → Na2CO3(aq) + Na2S(aq) + H2O(l) a. Balance the chemical equation. b. Identify the oxidant and the reductant. c. How much carbon is needed to convert 2.8 kg of sodium sulfate to sodium sulfide? d. If the yield of the reaction were only 78%, how many kilograms of sodium carbonate would be produced from 2.80 kg of sodium sulfate?
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e. If 240 g of carbon and 2.80 kg of sodium sulfate were used in the reaction, what would be the limiting reactant (assuming an excess of sodium hydroxide)? 14. The reaction of A2 (blue) with B2 (yellow) is shown below. The initial reaction mixture is shown on the left and the mixture after the reaction has gone to completion is shown on the right.
a. Write a balanced chemical equation for the reaction. b. Which is the limiting reactant in the initial reaction mixture? c. How many moles of the product AB4 could you obtain from a mixture of 0.020 mol A2 and 0.060 mol B2? 15. The reaction of X4 (orange) with Y2 (black) is shown below. The initial reaction mixture is shown on the left and the mixture after the reaction has gone to completion is shown on the right.
a. Write a balanced chemical equation for the reaction. b. Which is the limiting reactant in the initial reaction mixture? c. How many moles of the product XY3 could you obtain from a mixture of 0.100 mol X4 and 0.300 mol Y2?
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16. Methyl butyrate, an artificial apple flavor used in the food industry, is produced by the reaction of butanoic acid with methanol in the presence of an acid catalyst (H+): H+
CH 3 CH 2 CH 2 COOH(l) + CH 3 OH(l) ⎯→ CH 3 CH 2 CH 2 CO 2 CH 3 (l) + a. Given 7.8 g of butanoic acid, how many grams of methyl butyrate would be synthesized, assuming 100% yield? b. The reaction produced 5.5 g of methyl butyrate. What was the percent yield? c. Is the catalyst used in this reaction heterogeneous or homogeneous? 17. In the presence of a platinum catalyst, hydrogen and bromine react at elevated temperatures (300°C) to form hydrogen bromide (heat is indicated by Δ): Pt
H 2 (g) + Br2 (l) ⎯→ 2HBr(g) Δ Given the following, calculate the mass of hydrogen bromide produced: a. b. c. d.
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8.23 × 1022 molecules of H2 6.1 × 103 mol of H2 1.3 × 105 g of H2 Is the catalyst used in this reaction heterogeneous or homogeneous?
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ANSWERS 1. a. b. c. d.
redox reaction exchange acid–base condensation
3. a. b. c. d. e. f. g. h.
S, −2; N, −3; H, +1 P, +5; O, −2 F, −1; Al, +3 S, −2; Cu, +2 H, +1; O, −2; C, +4 H, +1; N, −3 H, +1; O, −2; S, +6 H, +1, O, −2; C, +2
i. butanol:
O,−2; H, +1 From left to right: C, −3–2–2–1 5. 2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l) acid–base reaction 7. a. b. c. d. e. f. g. h.
Ca, +2; O, −2; C, +4 Na, +1; Cl, −1 O, −2; C, +4 K, +1; O, −2; Cr, +6 K, +1; O, −2; Mn, +7 O, −2; Fe, +3 O, −2; H, +1; Cu, +2 O, −2; S, +6
i. Hexanol
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O,−2; H, +1 From left to right: C: −3, −2, −2, −2, −2, −1 9. a. Na is the reductant and is oxidized. Cl2 is the oxidant and is reduced. b. Mg is the reductant and is oxidized. Si is the oxidant and is reduced. c. H2O2 is both the oxidant and reductant. One molecule is oxidized, and one molecule is reduced. Δ
11. a. H2O(g) + C(s) ⎯→ H2(g) + CO(g) C is the reductant and is oxidized. H2O is the oxidant and is reduced. b. 8Mn(s) + S8(s) + 8CaO(s) → 8CaS(s) + 8MnO(s) Mn is the reductant and is oxidized. The S8 is the oxidant and is reduced. Δ
c. 2C2H4(g) + O2(g) ⎯→ 2C2H4O(g) Ethylene is the reductant and is oxidized. O2 is the oxidant and is reduced. d. 8ZnS(s) + 8H2SO4(aq) + 4O2(g) → 8ZnSO4(aq) + S8(s) + 8H2O(l) Sulfide in ZnS is the reductant and is oxidized. O2 is the oxidant and is reduced. 13. a. b. c. d. e.
Na2SO4 + 2C + 4NaOH → 2Na2CO3 + Na2S + 2H2O The sulfate ion is the oxidant, and the reductant is carbon. 470 g 3300 g carbon
17. a. b. c. d.
22.1 g 9.9 × 105 g 1.0 × 107 g heterogeneous
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3.6 Chemical Reactions in the Atmosphere LEARNING OBJECTIVE 1. To become familiar with the various reactions implicated in the destruction of Earth’s ozone layer.
Section 3.5 "Classifying Chemical Reactions" described different classes of chemical reactions. Of the many different chemical reactions that occur in Earth’s atmosphere, some are important and controversial because they affect our quality of life and health. The atmospheric reactions presented in this section provide examples of the various classes of reactions introduced in this chapter that are implicated in the destruction of Earth’s protective ozone layer37.
37. A concentration of ozone in the stratosphere (about 1015 ozone molecules per liter) that acts as a protective screen, absorbing ultraviolet light that would otherwise reach the surface of the earth, where it would harm plants and animals. 38. An unstable form of oxygen that consists of three oxygen atoms bonded together (O3). A layer of ozone in the stratosphere helps protect the plants and animals on earth from harmful ultraviolet radiation. Ozone is responsible for the pungent smell we associate with lightning discharges and electric motors. It is also toxic.
Each year since the mid-1970s, scientists have noted a disappearance of approximately 70% of the ozone (O3) layer above Antarctica during the Antarctic spring, creating what is commonly known as the “ozone hole.” Ozone38 is an unstable form of oxygen that consists of three oxygen atoms bonded together. In September 2009, the Antarctic ozone hole reached 24.1 million km2 (9.3 million mi2), about the size of North America. The largest area ever recorded was in the year 2000, when the hole measured 29.9 million km2 and for the first time extended over a populated area—the city of Punta Arenas, Chile (population 154,000; Figure 3.15 "Satellite Photos of Earth Reveal the Sizes of the Antarctic Ozone Hole over Time"). A less extensive zone of depletion has been detected over the Arctic as well. Years of study from the ground, from the air, and from satellites in space have shown that chlorine from industrial chemicals used in spray cans, foam packaging, and refrigeration materials is largely responsible for the catalytic depletion of ozone through a series of condensation, cleavage, and oxidation–reduction reactions.
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Figure 3.15 Satellite Photos of Earth Reveal the Sizes of the Antarctic Ozone Hole over Time
Dark blue colors correspond to the thinnest ozone; light blue, green, yellow, orange, and red indicate progressively thicker ozone. In September 2000, the Antarctic ozone hole briefly approached a record 30 million km 2.
Earth’s Atmosphere and the Ozone Layer Earth’s atmosphere at sea level is an approximately 80:20 solution of nitrogen and oxygen gases, with small amounts of carbon dioxide, water vapor, and the noble gases, and trace amounts of a variety of other compounds (Table 3.2 "The Composition of Earth’s Atmosphere at Sea Level*"). A key feature of the atmosphere is that its composition, temperature, and pressure vary dramatically with altitude. Consequently, scientists have divided the atmosphere into distinct layers, which interact differently with the continuous flux of solar radiation from the top and the land and ocean masses at the bottom. Some of the characteristic features of the layers of the atmosphere are illustrated in Figure 3.16 "Variation of Temperature with Altitude in Earth’s Atmosphere". Table 3.2 The Composition of Earth’s Atmosphere at Sea Level* Gas
Formula
Volume (%)
nitrogen
N2
78.084
oxygen
O2
20.948
argon
Ar
0.934
carbon dioxide†
CO2
0.0314
neon
Ne
0.00182
* In addition, air contains as much as 7% water vapor (H 2O), 0.0001% sulfur dioxide (SO2), 0.00007% ozone (O3), 0.000002% carbon monoxide (CO), and 0.000002% nitrogen dioxide (NO2). †
Carbon dioxide levels are highly variable; the typical range is 0.01–0.1%.
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Gas
Formula
Volume (%)
helium
He
0.000524
krypton
Kr
0.000114
methane
CH4
0.0002
hydrogen
H2
0.00005
nitrous oxide
N 2O
0.00005
xenon
Xe
0.0000087
* In addition, air contains as much as 7% water vapor (H 2O), 0.0001% sulfur dioxide (SO2), 0.00007% ozone (O3), 0.000002% carbon monoxide (CO), and 0.000002% nitrogen dioxide (NO2). †
Carbon dioxide levels are highly variable; the typical range is 0.01–0.1%.
Figure 3.16 Variation of Temperature with Altitude in Earth’s Atmosphere
Note the important chemical species present in each layer. The yellow line indicates the temperature at various altitudes.
39. The lowest layer of the atmosphere, the troposphere extends from earth’s surface to an altitude of about 11–13 km (7–8 miles). The temperature of the troposphere decreases steadily with increasing altitude.
The troposphere39 is the lowest layer of the atmosphere, extending from Earth’s surface to an altitude of about 11–13 km (7–8 mi). Above the troposphere lies the stratosphere, which extends from 13 km (8 mi) to about 44 km (27 mi). As shown in Figure 3.16 "Variation of Temperature with Altitude in Earth’s Atmosphere", the
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temperature of the troposphere decreases steadily with increasing altitude. Because “hot air rises,” this temperature gradient leads to continuous mixing of the upper and lower regions within the layer. The thermally induced turbulence in the troposphere produces fluctuations in temperature and precipitation that we collectively refer to as “weather.” In contrast, mixing between the layers of the atmosphere occurs relatively slowly, so each layer has distinctive chemistry. We focus our attention on the stratosphere, which contains the highest concentration of ozone. The sun’s radiation is the major source of energy that initiates chemical reactions in the atmosphere. The sun emits many kinds of radiation, including visible light40, which is radiation that the human eye can detect, and ultraviolet light41, which is higher energy radiation that cannot be detected by the human eye. This higher energy ultraviolet light can cause a wide variety of chemical reactions that are harmful to organisms. For example, ultraviolet light is used to sterilize items, and, as anyone who has ever suffered a severe sunburn knows, it can produce extensive tissue damage. Light in the higher energy ultraviolet range is almost totally absorbed by oxygen molecules in the upper layers of the atmosphere, causing the O 2 molecules to dissociate into two oxygen atoms in a cleavage reaction: Equation 3.34 light
O2 (g) ⎯⎯⎯→ 2O(g) In Equation 3.34, light is written above the arrow to indicate that light is required for the reaction to occur. The oxygen atoms produced in Equation 3.34 can undergo a condensation reaction with O2 molecules to form ozone: Equation 3.35 O(g) + O2(g) → O3(g) 40. Radiation that the human eye can detect. 41. High-energy radiation that cannot be detected by the human eye but can cause a wide variety of chemical reactions that are harmful to organisms.
Ozone is responsible for the pungent smell we associate with lightning discharges and electric motors. It is also toxic and a significant air pollutant, particularly in cities.
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In the stratosphere, the ozone produced via Equation 3.35 has a major beneficial effect. Ozone absorbs the less-energetic range of ultraviolet light, undergoing a cleavage reaction in the process to give O2 and O: Equation 3.36 light
O3 (g) ⎯⎯⎯→ O2 (g) + O(g) The formation of ozone (Equation 3.35) and its decomposition (Equation 3.36) are normally in balance, resulting in essentially constant levels of about 10 15 ozone molecules per liter in the stratosphere. This so-called ozone layer acts as a protective screen that absorbs ultraviolet light that would otherwise reach Earth’s surface. In 1974, F. Sherwood Rowland and Mario Molina published a paper claiming that commonly used chlorofluorocarbon (CFC) compounds were causing major damage to the ozone layer (Table 3.3 "Common CFCs and Related Compounds"). CFCs had been used as refrigerants and propellants in aerosol cans for many years, releasing millions of tons of CFC molecules into the atmosphere. Because CFCs are volatile compounds that do not readily undergo chemical reactions, they persist in the atmosphere long enough to be carried to the top of the troposphere, where they eventually enter the stratosphere. There they are exposed to intense ultraviolet light and undergo a cleavage reaction to produce a chlorine atom, which is shown for Freon-11: Equation 3.37 light
CCl3 F(g) ⎯⎯⎯→ CCl2 F(g) + Cl(g) The resulting chlorine atoms act as a homogeneous catalyst in two redox reactions (Equation 3.38 and Equation 3.39): Equation 3.38 Cl(g) + O3(g) → ClO(g) + O2(g) Equation 3.39 ClO(g) + O(g) → Cl(g) + O2(g)
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Adding the two reactions in Equation 3.38 and Equation 3.39 gives Equation 3.40 Cl(g) + O3(g) + ClO(g) + O(g) → ClO(g) + Cl(g) + 2O2(g) Because chlorine and ClO (chlorine monoxide) appear on both sides of the equation, they can be canceled to give the following net reaction: Equation 3.41 O3(g) + O(g) → 2O2(g) In the presence of chlorine atoms, one O3 molecule and one oxygen atom react to give two O2 molecules. Although chlorine is necessary for the overall reaction to occur, it does not appear in the net equation. The chlorine atoms are a catalyst that increases the rate at which ozone is converted to oxygen. Table 3.3 Common CFCs and Related Compounds Name
Molecular Formula
Industrial Name
trichlorofluoromethane
CCl3F
CFC-11 (Freon-11)
dichlorodifluoromethane
CCl2F2
CFC-12 (Freon-12)
chlorotrifluoromethane
CClF3
CFC-13 (Freon-13)
bromotrifluoromethane
CBrF3
Halon-1301*
bromochlorodifluoromethane
CBrClF2
Halon-1211
*Halons, compounds similar to CFCs that contain at least one bromine atom, are used as fire extinguishers in specific applications (e.g., the engine rooms of ships).
Because the stratosphere is relatively isolated from the layers of the atmosphere above and below it, once chlorine-containing species enter the stratosphere, they remain there for long periods of time. Each chlorine atom produced from a CFC molecule can lead to the destruction of large numbers of ozone molecules, thereby decreasing the concentration of ozone in the stratosphere. Eventually, however, the chlorine atom reacts with a water molecule to form hydrochloric acid, which is carried back into the troposphere and then washed out of the atmosphere in rainfall.
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The Ozone Hole Massive ozone depletions were first observed in 1975 over the Antarctic and more recently over the Arctic. Although the reactions in Equation 3.38 and Equation 3.39 appear to account for most of the ozone destruction observed at low to middle latitudes, Equation 3.37 requires intense sunlight to generate chlorine atoms, and sunlight is in very short supply during the polar winters. At high latitudes (near the poles), therefore, a different set of reactions must be responsible for the depletion. Recent research has shown that, in the absence of oxygen atoms, chlorine monoxide can react with stratospheric nitrogen dioxide in a redox reaction to form chlorine nitrate (ClONO2). When chlorine nitrate is in the presence of trace amounts of HCl or adsorbed on ice particles in stratospheric clouds, additional redox reactions can occur in which chlorine nitrate produces Cl2 or HOCl (hypochlorous acid): Equation 3.42 HCl(g) + ClONO2(g) → Cl2(g) + HNO3(g) Equation 3.43 H2O(g) + ClONO2(g) → HOCl(g) + HNO3(g) Both Cl2 and HOCl undergo cleavage reactions by even weak sunlight to give reactive chlorine atoms. When the sun finally rises after the long polar night, relatively large amounts of Cl2 and HOCl are present and rapidly generate high levels of chlorine atoms. The reactions shown in Equation 3.38 and Equation 3.39 then cause ozone levels to fall dramatically. Stratospheric ozone levels decreased about 2.5% from 1978 to 1988, which coincided with a fivefold increase in the widespread use of CFCs since the 1950s. If the trend were allowed to continue, the results could be catastrophic. Fortunately, many countries have banned the use of CFCs in aerosols. In 1987, representatives from 43 nations signed the Montreal Protocol, committing themselves to reducing CFC emissions by 50% by the year 2000. Later, representatives from a large number of countries, alarmed by data showing the rapid depletion of stratospheric chlorine, agreed to phase out CFCs completely by the early 21st century; the United States banned their use in 1995. The projected effects of these agreements on atmospheric chlorine levels are shown in Figure 3.17 "Projected Effects of International Agreements on Atmospheric Chlorine Levels". Because of the very slow rate at
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which CFCs are removed from the stratosphere, however, stratospheric chlorine levels will not fall to the level at which the Antarctic ozone hole was first observed until about 2050. The scientific community recognized Molina and Rowland’s work in 1995, when they shared the Nobel Prize in Chemistry. Figure 3.17 Projected Effects of International Agreements on Atmospheric Chlorine Levels
The graph plots atmospheric chlorine content in chlorine atoms per 10 9 molecules of O2 plus N2 from 1960 to 1990 (actual data) and 1990 to 2080 (estimated for various schemes for regulating CFC emissions).
Manufacturing companies are now under great political and economic pressure to find alternatives to the CFCs used in the air-conditioning units of cars, houses, and commercial buildings. One approach is to use hydrochlorofluorocarbons (HCFCs), hydrocarbons in which only some of the hydrogen atoms are replaced by chlorine or fluorine, and hydrofluorocarbons (HFCs), which do not contain chlorine (Table 3.4 "Selected HCFCs and HFCs"). The C–H bonds in HCFCs and HFCs act as “handles” that permit additional chemical reactions to occur. Consequently, these substances are degraded more rapidly, and most are washed out of the atmosphere before they can reach the stratosphere.
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Table 3.4 Selected HCFCs and HFCs Name
Molecular Formula
Industrial Name
chlorodifluoromethane
CHClF2
HCFC-22 (freon-22)
1-chloro-1,1-difluoroethane
CH3CClF2
HCFC-141b
2,2-dichloro-1,1,1-trifluoroethane CHCl2CF3 1,1,1,2-tetrafluoroethane
CH2FCF3
HCFC-123 HFC-134a
Nonetheless, the small fraction of HCFCs that reaches the stratosphere will deplete ozone levels just as CFCs do, so they are not the final answer. Indeed, the 1990 London amendment to the Montreal Protocol specifies that HCFCs must be phased out by 2040. Finding a suitable replacement for refrigerants is just one of the challenges facing chemists in the 21st century.
HFCs are used as a replacement for CFCs. The molecular structure of HFC-134a is shown in this ball-and-stick model.
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EXAMPLE 16 Nitric oxide (NO) may also be an important factor in the destruction of the ozone layer. One source of this compound is the combustion of hydrocarbons in jet engines. The fact that high-flying supersonic aircraft inject NO directly into the stratosphere was a major argument against the development of commercial supersonic transports. Do you agree with this decision? Why or why not? Given: identity of compound Asked for: assessment of likely role in ozone depletion Strategy: Predict what reactions are likely to occur between NO and ozone and then determine whether the reactions are likely to deplete ozone from the atmosphere. Solution: Both NO and NO2 are known oxides of nitrogen. Thus NO is likely to react with ozone according to the chemical equation NO(g) + O3(g) → NO2(g) + O2(g) resulting in ozone depletion. If NO2(g) also reacts with atomic oxygen according to the equation NO2(g) + O(g) → NO(g) + O2(g) then we would have a potential catalytic cycle for ozone destruction similar to that caused by chlorine atoms. Based on these reactions, the development of commercial supersonic transports is not recommended until the environmental impact has undergone additional testing. (Although these reactions have been observed, they do not appear to be a major factor in ozone destruction.) Exercise
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An industrial manufacturer proposed that halons such as CF 3Br could be used as replacements for CFC propellants. Do you think that this is a reasonable suggestion or is there a potential problem with such a use?
Answer: Because the compound CF3Br contains carbon, fluorine, and a bromine atom that is chemically similar to chlorine, it is likely that it would also be a catalyst for ozone destruction. There is therefore a potential problem with its use.
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Summary Earth’s atmosphere consists of discrete layers that do not mix readily with one another. The sun emits radiation with a wide range of energies, including visible light, which can be detected by the human eye, and ultraviolet light, which is more energetic than visible light and cannot be detected by the human eye. In the stratosphere, ultraviolet light reacts with O2 molecules to form atomic oxygen. Atomic oxygen then reacts with an O2 molecule to produce ozone (O3). As a result of this reaction, the stratosphere contains an appreciable concentration of ozone molecules that constitutes the ozone layer. The absorption of ultraviolet light in the stratosphere protects Earth’s surface from the sun’s harmful effects. Volatile organic compounds that contain chlorine and fluorine, which are known as chlorofluorocarbons (CFCs), are capable of reaching the stratosphere, where they can react with ultraviolet light to generate chlorine atoms and other chlorine-containing species that catalyze the conversion of ozone to O2, thereby decreasing the amount of O3 in the stratosphere. Replacing chlorofluorocarbons with hydrochlorofluorocarbons (HCFCs) or hydrofluorocarbons (HFCs) is one strategy that has been developed to minimize further damage to Earth’s ozone layer.
KEY TAKEAWAY • The composition of Earth’s atmosphere is vulnerable to degradation through reactions with common industrial chemicals.
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CONCEPTUAL PROBLEMS 1. Carbon monoxide is a toxic gas that can be produced from the combustion of wood in wood-burning stoves when excess oxygen is not present. Write a balanced chemical equation showing how carbon monoxide is produced from carbon and suggest what might be done to prevent it from being a reaction product. 2. Explain why stratospheric ozone depletion has developed over the coldest part of Earth (the poles) and reaches a maximum at the beginning of the polar spring. 3. What type of reactions produce species that are believed to be responsible for catalytic depletion of ozone in the atmosphere?
NUMERICAL PROBLEM Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems. 1. Sulfur dioxide and hydrogen sulfide are important atmospheric contaminants that have resulted in the deterioration of ancient objects. Sulfur dioxide combines with water to produce sulfurous acid, which then reacts with atmospheric oxygen to produce sulfuric acid. Sulfuric acid is known to attack many metals that were used by ancient cultures. Give the formulas for these four sulfur-containing species. What is the percentage of sulfur in each compound? What is the percentage of oxygen in each?
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3.7 Essential Skills 2 TOPICS • Proportions • Percentages • Unit Conversions
In Essential Skills 1 in Chapter 1 "Introduction to Chemistry", Section 1.9 "Essential Skills 1", we introduced you to some of the fundamental mathematical operations you need to successfully manipulate mathematical equations in chemistry. Before proceeding to the problems in Chapter 3 "Chemical Reactions", you should become familiar with the additional skills described in this section on proportions, percentages, and unit conversions.
Proportions We can solve many problems in general chemistry by using ratios, or proportions. For example, if the ratio of some quantity A to some quantity B is known, and the relationship between these quantities is known to be constant, then any change in A (from A1 to A2) produces a proportional change in B (from B1 to B2) and vice versa. The relationship between A1, B1, A2, and B2 can be written as follows:
A1 A2 = = constant B1 B2 To solve this equation for A2, we multiply both sides of the equality by B2, thus canceling B2 from the denominator:
(B 2 )
A1 A2 = ( B2 ) B1 B2 B2 A 1 = A2 B1
Similarly, we can solve for B2 by multiplying both sides of the equality by 1/A2, thus canceling A2 from the numerator:
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1 A1 1 = ( A 2 ) B1 A2 A1 1 = A 2 B1 B2
A2 B 2
If the values of A1, A2, and B1 are known, then we can solve the left side of the equation and invert the answer to obtain B2:
1 = numerical value B2 1 B2 = numerical value If the value of A1, A2, or B1 is unknown, however, we can solve for B2 by inverting both sides of the equality:
B2 =
A 2 B1 A1
When you manipulate equations, remember that any operation carried out on one side of the equality must be carried out on the other. Skill Builder ES1 illustrates how to find the value of an unknown by using proportions.
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SKILL BUILDER ES1 If 38.4 g of element A are needed to combine with 17.8 g of element B, then how many grams of element A are needed to combine with 52.3 g of element B? Solution We set up the proportions as follows:
A1
=
38.4 g
A2 B2 A1 B1 38.4 g 17.8 g
= =
? 52.3 g A2 B2 A2 52.3 g
B1
=
17.8 g
= =
Multiplying both sides of the equation by 52.3 g gives
(38.4 g)(52.3 g) 17.8 g
=
A2
=
A 2 (52.3 g) (52.3 g) 113 g
Notice that grams cancel to leave us with an answer that is in the correct units. Always check to make sure that your answer has the correct units.
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SKILL BUILDER ES2 Solve to find the indicated variable. a. b. c. d. e.
16.4 g x = 18.3 41.2 g g 2.65 m 3.28 m = y 4.02 m −3 3.27×10 g 5.0×10 −1 g = x 3.2 g P1 V2 Solve for V1: P = V 2 1 P1 V1 PV Solve for T1: T = T2 2 2 1
Solution
a. Multiply both sides of the equality by 18.3 g to remove this measurement from the denominator:
(18.3 g)
16.4 g 41.2 g
= (18.3 g)
x 18.3 g
7.28 g = x b. Multiply both sides of the equality by 1/3.28 m, solve the left side of the equation, and then invert to solve for y:
3.28 m 1 2.65 m 1 1 = = ( 3.28 m ) 4.02 m y y ( 3.28 m ) (4.02)(3.28) y= = 4.98 m 2.65 c. Multiply both sides of the equality by 1/3.27 × 10−3 g, solve the right side of the equation, and then invert to find x:
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1 −3 3.27 × 10 g
3.27 × 10 −3 g 5.0 × 1 1 = x 3.2 ( 3.27 × 10 −3 g ) x=
(3.2 g)(3.27 × 10 −3 g)
= 2.1 ×
5.0 × 10 −1 g
d. Multiply both sides of the equality by 1/V2, and then invert both sides to obtain V1:
1 P1 1 = ( V 2 ) P2 V2 P2 V 2 = V1 P1
V2 V 1
e. Multiply both sides of the equality by 1/P1V1 and then invert both sides to obtain T1:
1 P1 V 1
P1 V 1 1 P2 V 2 = T ( P1 V 1 ) T 2 1 1 P2 V 2 = T1 T 2 P1 V 1 T 2 P1 V 1 T1 = P2 V 2
Percentages Because many measurements are reported as percentages, many chemical calculations require an understanding of how to manipulate such values. You may, for example, need to calculate the mass percentage of a substance, as described in Section 3.2 "Determining Empirical and Molecular Formulas", or determine the percentage of product obtained from a particular reaction mixture. You can convert a percentage to decimal form by dividing the percentage by 100:
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52.8% =
52.8 = 0.528 100
Conversely, you can convert a decimal to a percentage by multiplying the decimal by 100: 0.356 × 100 = 35.6% Suppose, for example, you want to determine the mass of substance A, one component of a sample with a mass of 27 mg, and you are told that the sample consists of 82% A. You begin by converting the percentage to decimal form:
82% =
82 = 0.82 100
The mass of A can then be calculated from the mass of the sample: 0.82 × 27 mg = 22 mg Skill Builder ES3 provides practice in converting and using percentages.
SKILL BUILDER ES3 Convert each number to a percentage or a decimal. a. b. c. d.
29.4% 0.390 101% 1.023
Solution a.
29.4 100
= 0.294
101 100
= 1.01
b. 0.390 × 100 = 39.0% c.
d. 1.023 × 100 = 102.3%
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SKILL BUILDER ES4 Use percentages to answer the following questions, being sure to use the correct number of significant figures (see Essential Skills 1 in Chapter 1 "Introduction to Chemistry", Section 1.9 "Essential Skills 1"). Express your answer in scientific notation where appropriate. a. What is the mass of hydrogen in 52.83 g of a compound that is 11.2% hydrogen? b. What is the percentage of carbon in 28.4 g of a compound that contains 13.79 g of that element? c. A compound that is 4.08% oxygen contains 194 mg of that element. What is the mass of the compound? Solution
a. b.
52.83 g ×
11.2 100
13.79 g carbon 28.4 g
= 52.83 g × 0.112 = 5.92 g
× 100 = 48.6% carbon
c. This problem can be solved by using a proportion:
4.08% oxygen 100% compound
=
x
=
194 mg x mg
4.75 × 10 3 mg (or 4.75 g)
Unit Conversions As you learned in Essential Skills 1 in Chapter 1 "Introduction to Chemistry", Section 1.9 "Essential Skills 1", all measurements must be expressed in the correct units to have any meaning. This sometimes requires converting between different units (Table 1.7 "SI Base Units"). Conversions are carried out using conversion factors, which are are ratios constructed from the relationships between different units or measurements. The relationship between milligrams and grams, for example, can be expressed as either 1 g/1000 mg or 1000 mg/1 g. When making unit conversions, use arithmetic steps accompanied by unit cancellation.
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Suppose you have measured a mass in milligrams but need to report the measurement in kilograms. In problems that involve SI units, you can use the definitions of the prefixes given in Table 1.6 "Approximate Elemental Composition of a Typical 70 kg Human" to get the necessary conversion factors. For example, you can convert milligrams to grams and then convert grams to kilograms: milligrams → grams → kilograms 1000 mg → 1 g 1000 g → 1 kilogram If you have measured 928 mg of a substance, you can convert that value to kilograms as follows:
928 mg × 0.928 g ×
1g = 0.928 g 1000 mg
1 kg = 0.000928 kg = 9.28 × 10 −4 kg 1000 g
In each arithmetic step, the units cancel as if they were algebraic variables, leaving us with an answer in kilograms. In the conversion to grams, we begin with milligrams in the numerator. Milligrams must therefore appear in the denominator of the conversion factor to produce an answer in grams. The individual steps may be connected as follows:
928 mg ×
1 g 1000 mg
×
1 kg 928 kg = = 928 × 10 −6 kg = 9.28 6 1000 g 10
Skill Builder ES5 provides practice converting between units.
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SKILL BUILDER ES5 Use the information in Table 1.8 "Prefixes Used with SI Units" to convert each measurement. Be sure that your answers contain the correct number of significant figures and are expressed in scientific notation where appropriate. a. b. c. d. e.
59.2 cm to decimeters 3.7 × 105 mg to kilograms 270 mL to cubic decimeters 2.04 × 103 g to tons 9.024 × 1010 s to years
Solution
3.7 Essential Skills 2
1 m
a.
59.2 cm ×
b.
3.7 × 10 5 mg ×
c.
270 mL ×
d.
2.04 × 10 3 g ×
e.
9.024 × 10 10 s ×
100 cm
10 dm 1 m 1 g
×
×
1000 mg
1 L
×
1000 mL 1 lb
453.6 g 1 min 60 s
= 5.92 dm
1 dm 3 1 L
× ×
1 kg 1000 g
= 3.7 × 10 −1 kg
= 270 × 10 −3 dm 3 = 2.70 × 10 −1 dm
1 tn 2000 lb 1 h 60 min
= 0.00225 tn = 2.25 × 10 −3 tn ×
1 d 24 h
×
1 yr 365 d
= 2.86 × 10 3
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APPLICATION PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts. 1. Hydrogen sulfide is a noxious and toxic gas produced from decaying organic matter that contains sulfur. A lethal concentration in rats corresponds to an inhaled dose of 715 molecules per million molecules of air. How many molecules does this correspond to per mole of air? How many moles of hydrogen sulfide does this correspond to per mole of air? 2. Bromine, sometimes produced from brines (salt lakes) and ocean water, can be used for bleaching fibers and silks. How many moles of bromine atoms are found in 8.0 g of molecular bromine (Br2)? 3. Paris yellow is a lead compound that is used as a pigment; it contains 16.09% chromium, 19.80% oxygen, and 64.11% lead. What is the empirical formula of Paris yellow? 4. A particular chromium compound used for dyeing and waterproofing fabrics has the elemental composition 18.36% chromium, 13.81% potassium, 45.19% oxygen, and 22.64% sulfur. What is the empirical formula of this compound? 5. Compounds with aluminum and silicon are commonly found in the clay fractions of soils derived from volcanic ash. One of these compounds is vermiculite, which is formed in reactions caused by exposure to weather. Vermiculite has the following formula: Ca0.7[Si6.6Al1.4]Al4O20(OH)4. (The content of calcium, silicon, and aluminum are not shown as integers because the relative amounts of these elements vary from sample to sample.) What is the mass percent of each element in this sample of vermiculite? 6. ♦ Pheromones are chemical signals secreted by a member of one species to evoke a response in another member of the same species. One honeybee pheromone is an organic compound known as an alarm pheromone, which smells like bananas. It induces an aggressive attack by other honeybees, causing swarms of angry bees to attack the same aggressor. The composition of this alarm pheromone is 64.58% carbon, 10.84% hydrogen, and 24.58% oxygen by mass, and its molecular mass is 130.2 amu. a. Calculate the empirical formula of this pheromone. b. Determine its molecular formula. c. Assuming a honeybee secretes 1.00 × 10−11 g of pure pheromone, how many molecules of pheromone are secreted?
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7. Amoxicillin is a prescription drug used to treat a wide variety of bacterial infections, including infections of the middle ear and the upper and lower respiratory tracts. It destroys the cell walls of bacteria, which causes them to die. The elemental composition of amoxicillin is 52.59% carbon, 5.24% hydrogen, 11.50% nitrogen, 21.89% oxygen, and 8.77% sulfur by mass. What is its empirical formula? 8. Monosodium glutamate (MSG; molar mass = 169 g/mol), is used as a flavor enhancer in food preparation. It is known to cause headaches and chest pains in some individuals, the so-called Chinese food syndrome. Its composition was found to be 35.51% carbon, 4.77% hydrogen, 8.28% nitrogen, and 13.59% sodium by mass. If the “missing” mass is oxygen, what is the empirical formula of MSG? 9. Ritalin is a mild central nervous system stimulant that is prescribed to treat attention deficit disorders and narcolepsy (an uncontrollable desire to sleep). Its chemical name is methylphenidate hydrochloride, and its empirical formula is C14H20ClNO2. If you sent a sample of this compound to a commercial laboratory for elemental analysis, what results would you expect for the mass percentages of carbon, hydrogen, and nitrogen? 10. Fructose, a sugar found in fruit, contains only carbon, oxygen, and hydrogen. It is used in ice cream to prevent a sandy texture. Complete combustion of 32.4 mg of fructose in oxygen produced 47.6 mg of CO2 and 19.4 mg of H2O. What is the empirical formula of fructose? 11. Coniine, the primary toxin in hemlock, contains only carbon, nitrogen, and hydrogen. When ingested, it causes paralysis and eventual death. Complete combustion of 28.7 mg of coniine produced 79.4 mg of CO 2 and 34.4 mg of H2O. What is the empirical formula of the coniine? 12. Copper and tin alloys (bronzes) with a high arsenic content were presumably used by Bronze Age metallurgists because bronze produced from arsenic-rich ores had superior casting and working properties. The compositions of some representative bronzes of this type are as follows: Origin
Dead Sea
% Composition Cu
As
87.0
12.0
Central America 90.7
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If ancient metallurgists had used the mineral As2S3 as their source of arsenic, how much As2S3 would have been required to process 100 g of cuprite (Cu2O) bronzes with these compositions? 13. ♦ The phrase mad as a hatter refers to mental disorders caused by exposure to mercury(II) nitrate in the felt hat manufacturing trade during the 18th and 19th centuries. An even greater danger to humans, however, arises from alkyl derivatives of mercury. a. Give the empirical formula of mercury(II) nitrate. b. One alkyl derivative, dimethylmercury, is a highly toxic compound that can cause mercury poisoning in humans. How many molecules are contained in a 5.0 g sample of dimethylmercury? c. What is the percentage of mercury in the sample? 14. Magnesium carbonate, aluminum hydroxide, and sodium bicarbonate are commonly used as antacids. Give the empirical formulas and determine the molar masses of these compounds. Based on their formulas, suggest another compound that might be an effective antacid. 15. ♦ Nickel(II) acetate, lead(II) phosphate, zinc nitrate, and beryllium oxide have all been reported to induce cancers in experimental animals. a. Give the empirical formulas for these compounds. b. Calculate their formula masses. c. Based on the location of cadmium in the periodic table, would you predict that cadmium chloride might also induce cancer? 16. ♦ Methane, the major component of natural gas, is found in the atmospheres of Jupiter, Saturn, Uranus, and Neptune. a. What is the structure of methane? b. Calculate the molecular mass of methane. c. Calculate the mass percentage of both elements present in methane. 17. Sodium saccharin, which is approximately 500 times sweeter than sucrose, is frequently used as a sugar substitute. What are the percentages of carbon, oxygen, and sulfur in this artificial sweetener?
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18. Lactic acid, found in sour milk, dill pickles, and sauerkraut, has the functional groups of both an alcohol and a carboxylic acid. The empirical formula for this compound is CH2O, and its molar mass is 90 g/mol. If this compound were sent to a laboratory for elemental analysis, what results would you expect for carbon, hydrogen, and oxygen content? 19. The compound 2-nonenal is a cockroach repellant that is found in cucumbers, watermelon, and carrots. Determine its molecular mass.
20. You have obtained a 720 mg sample of what you believe to be pure fructose, although it is possible that the sample has been contaminated with formaldehyde. Fructose and formaldehyde both have the empirical formula CH2O. Could you use the results from combustion analysis to determine whether your sample is pure? 21. ♦ The booster rockets in the space shuttles used a mixture of aluminum metal and ammonium perchlorate for fuel. Upon ignition, this mixture can react according to the chemical equation Al(s) + NH4ClO4(s) → Al2O3(s) + AlCl3(g) + NO(g) + H2O(g) Balance the equation and construct a table showing how to interpret this information in terms of the following: a. b. c. d.
numbers of individual atoms, molecules, and ions moles of reactants and products grams of reactants and products numbers of molecules of reactants and products given 1 mol of aluminum metal
22. ♦ One of the byproducts of the manufacturing of soap is glycerol. In 1847, it was discovered that the reaction of glycerol with nitric acid produced nitroglycerin according to the following unbalanced chemical equation:
Nitroglycerine is both an explosive liquid and a blood vessel dilator that is used to treat a heart condition known as angina.
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a. Balance the chemical equation and determine how many grams of nitroglycerine would be produced from 15.00 g of glycerol. b. If 9.00 g of nitric acid had been used in the reaction, which would be the limiting reactant? c. What is the theoretical yield in grams of nitroglycerin? d. If 9.3 g of nitroglycerin was produced from 9.0 g of nitric acid, what would be the percent yield? e. Given the data in part d, how would you rate the success of this reaction according to the criteria mentioned in this chapter? f. Derive a general expression for the theoretical yield of nitroglycerin in terms of x grams of glycerol. 23. ♦ A significant weathering reaction in geochemistry is hydration–dehydration. An example is the transformation of hematite (Fe2O3) to ferrihydrite (Fe10O15·9H2O) as the relative humidity of the soil approaches 100%: Fe2O3(s) + H2O(l) → Fe10O15·9H2O(s) This reaction occurs during advanced stages of the weathering process. a. Balance the chemical equation. b. Is this a redox reaction? Explain your answer. c. If 1 tn of hematite rock weathered in this manner, how many kilograms of ferrihydrite would be formed? 24. ♦ Hydrazine (N2H4) is used not only as a rocket fuel but also in industry to remove toxic chromates from waste water according to the following chemical equation: 4CrO42−(aq) + 3N2H4(l) + 4H2O(l) → 4Cr(OH)3(s) + 3N2(g) + 8OH−(aq) Identify the species that is oxidized and the species that is reduced. What mass of water is needed for the complete reaction of 15.0 kg of hydrazine? Write a general equation for the mass of chromium(III) hydroxide [Cr(OH) 3] produced from x grams of hydrazine. 25. ♦ Corrosion is a term for the deterioration of metals through chemical reaction with their environment. A particularly difficult problem for the archaeological chemist is the formation of CuCl, an unstable compound that is formed by the corrosion of copper and its alloys. Although copper and bronze objects can survive burial for centuries without significant deterioration, exposure to air can cause cuprous chloride to react with atmospheric oxygen to form Cu 2O and cupric chloride. The cupric chloride then reacts with the free metal to produce cuprous chloride. Continued reaction of oxygen and water with cuprous chloride causes “bronze disease,” which consists of spots of a pale green, powdery deposit of [CuCl2·3Cu(OH)2·H2O] on the surface of the object
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that continues to grow. Using this series of reactions described, complete and balance the following equations, which together result in bronze disease: Equation 1: ___ + O2 → ___ + ___ Equation 2: ___ + Cu → ___ Equation 3: ___ + O2 + H2O → CuCl 2
· 3Cu(OH) 2 · H 2 O + CuCl 2 bronze disease
a. Which species are the oxidants and the reductants in each equation? b. If 8.0% by mass of a 350.0 kg copper statue consisted of CuCl, and the statue succumbed to bronze disease, how many pounds of the powdery green hydrate would be formed? c. What factors could affect the rate of deterioration of a recently excavated bronze artifact? 26. ♦ Iron submerged in seawater will react with dissolved oxygen, but when an iron object, such as a ship, sinks into the seabed where there is little or no free oxygen, the iron remains fresh until it is brought to the surface. Even in the seabed, however, iron can react with salt water according to the following unbalanced chemical equation: Fe(s) + NaCl(aq) + H2O(l) → FeCl2(s) + NaOH(aq) + H2(g) The ferrous chloride and water then form hydrated ferrous chloride according to the following equation: FeCl2(s) + 2H2O(l) → FeCl2·2H2O(s) When the submerged iron object is removed from the seabed, the ferrous chloride dihydrate reacts with atmospheric moisture to form a solution that seeps outward, producing a characteristic “sweat” that may continue to emerge for many years. Oxygen from the air oxidizes the solution to ferric resulting in the formation of what is commonly referred to as rust (ferric oxide): FeCl2(aq) + O2(g) → FeCl3(aq) + Fe2O3(s) The rust layer will continue to grow until arrested. a. Balance each chemical equation. b. Given a 10.0 tn ship of which 2.60% is now rust, how many kilograms of iron were converted to FeCl2, assuming that the ship was pure iron? c. What mass of rust in grams would result? d. What is the overall change in the oxidation state of iron for this process? e. In the first equation given, what species has been reduced? What species has been oxidized?
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27. ♦ The glass industry uses lead oxide in the production of fine crystal glass, such as crystal goblets. Lead oxide can be formed by the following reaction: PbS(s) + O2(g) → PbO(s) + SO2(g) Balance the equation and determine what has been oxidized and what has been reduced. How many grams of sulfur dioxide would be produced from 4.0 × 103 g of lead sulfide? Discuss some potential environmental hazards that stem from this reaction. 28. ♦ The Deacon process is one way to recover Cl2 on-site in industrial plants where the chlorination of hydrocarbons produces HCl. The reaction uses oxygen to oxidize HCl to chlorine, as shown. HCl(g) + O2(g) → Cl2(g) + H2O(g) The reaction is frequently carried out in the presence of NO as a catalyst. a. Balance the chemical equation. b. Which compound is the oxidant, and which is the reductant? c. If 26 kg of HCl was produced during a chlorination reaction, how many kilograms of water would result from the Deacon process? 29. In 1834, Eilhardt Mitscherlich of the University of Berlin synthesized benzene (C6H6) by heating benzoic acid (C6H5COOH) with calcium oxide according to this balanced chemical equation: Δ
C 6 H 5 COOH(s) + CaO(s) ⎯→ C 6 H 6 (l) + CaCO 3 (s) (Heating is indicated by the symbol Δ.) How much benzene would you expect from the reaction of 16.9 g of benzoic acid and 18.4 g of calcium oxide? Which is the limiting reactant? How many grams of benzene would you expect to obtain from this reaction, assuming a 73% yield? 30. Aspirin (C9H8O4) is synthesized by the reaction of salicylic acid (C 7H6O3) with acetic anhydride (C4H6O3) according to the following equation: C7H6O3(s) + C4H6O3(l) → C9H8O4(s) + H2O(l) Balance the equation and find the limiting reactant given 10.0 g of acetic anhydride and 8.0 g of salicylic acid. How many grams of aspirin would you expect from this reaction, assuming an 83% yield? 31. ♦ Hydrofluoric acid etches glass because it dissolves silicon dioxide, as represented in the following chemical equation: SiO2(s) + HF(aq) → SiF62−(aq) + H+(aq) + H2O(l) a. Balance the equation. b. How many grams of silicon dioxide will react with 5.6 g of HF?
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c. How many grams of HF are needed to remove 80% of the silicon dioxide from a 4.0 kg piece of glass? (Assume that the glass is pure silicon dioxide.) 32. ♦ Lead sulfide and hydrogen peroxide react to form lead sulfate and water. This reaction is used to clean oil paintings that have blackened due to the reaction of the lead-based paints with atmospheric hydrogen sulfide. a. Write the balanced chemical equation for the oxidation of lead sulfide by hydrogen peroxide. b. What mass of hydrogen peroxide would be needed to remove 3.4 g of lead sulfide? c. If the painting had originally been covered with 5.4 g of lead sulfide and you had 3.0 g of hydrogen peroxide, what percent of the lead sulfide could be removed? 33. ♦ It has been suggested that diacetylene (C4H2, HC≡C–C≡CH) may be the ozone of the outer planets. As the largest hydrocarbon yet identified in planetary atmospheres, diacetylene shields planetary surfaces from ultraviolet radiation and is itself reactive when exposed to light. One reaction of diacetylene is an important route for the formation of higher hydrocarbons, as shown in the following chemical equations: C4H2(g) + C4H2(g) → C8H3(g) + H(g) C8H3(g) + C4H2(g) → C10H3(g) + C2H2(g) Consider the second reaction. a. Given 18.4 mol of C8H3 and 1000 g of C4H2, which is the limiting reactant? b. Given 2.8 × 1024 molecules of C8H3 and 250 g of C4H2, which is the limiting reactant? c. Given 385 g of C8H3 and 200 g of C4H2, which is in excess? How many grams of excess reactant would remain? d. Suggest why this reaction might be of interest to scientists. 34. ♦ Glucose (C6H12O6) can be converted to ethanol and carbon dioxide using certain enzymes. As alcohol concentrations are increased, however, catalytic activity is inhibited, and alcohol production ceases. a. Write a balanced chemical equation for the conversion of glucose to ethanol and carbon dioxide. b. Given 12.6 g of glucose, how many grams of ethanol would be produced, assuming complete conversion? c. If 4.3 g of ethanol had been produced, what would be the percent yield for this reaction? d. Is a heterogeneous catalyst or a homogeneous catalyst used in this reaction?
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e. You have been asked to find a way to increase the rate of this reaction given stoichiometric quantities of each reactant. How would you do this? 35. Early spacecraft developed by the National Aeronautics and Space Administration for its manned missions used capsules that had a pure oxygen atmosphere. This practice was stopped when a spark from an electrical short in the wiring inside the capsule of the Apollo 1 spacecraft ignited its contents. The resulting explosion and fire killed the three astronauts on board within minutes. What chemical steps could have been taken to prevent this disaster?
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ANSWERS 1. 4.31 × 1020 molecules, 7.15 × 10−4 3. PbCrO4 5. To two decimal places, the percentages are: H: 0.54%; O: 51.39%; Al: 19.50%; Si: 24.81%; Ca: 3.75% 7. C16H19O5N3S 13. a. Hg(NO3)2 b. 1.3 × 1022 molecules c. 86.96% mercury by mass. 15. a. Ni(O2CCH3)2; Pb3(PO4)2; Zn(NO3)2; BeO b. To four significant figures, the values are: Ni(O2CCH3)2, 176.8 amu; Pb3(PO4)2, 811.5 amu; Zn(NO3)2, 189.4 amu; BeO, 25.01 amu. c. Yes. 17. C, 40.98%; O, 23.39%; S, 15.63% 19. 140.22 amu 21. 3Al(s) + 3NH4ClO4(s) → Al2O3(s) + AlCl3(g) + 3NO(g) + 6H2O(g) 3NH4ClO4
3Al
Al2O3
a. 3 atoms 30 atoms, 6 ions 5 atoms b. 3 mol
3 mol
1 mol
c. 81 g
352 g
102 g
d. 6 × 1023 6 × 1023
2 × 1023
AlCl3
6 atoms, 3 molecules
18 atoms, 6 molecules
b. 1 mol
3 mol
6 mol
c. 133 g
90 g
108 g
d. 2 × 1023
6 × 1023
1.2 × 1022
a.
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4 atoms, 1 molecule
6H2O
3NO
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23. a. 5Fe2O3 + 9H2O → Fe10O15 · 9H2O b. No. c. 1090 kg 25. Equation 1: 8CuCl + O2 → 2Cu2O + 4CuCl2 Equation 2: CuCl2 + Cu → 2CuCl Equation 3: 12CuCl + 3O2 + 8H2O → 2[CuCl2 3Cu(OH)2 H2O] + 4CuCl2 a. Equation 1: Oxygen is the oxidant, and CuCl is the reductant. Equation 2: Copper is the reductant, and copper(II) chloride is the oxidant. Equation 3: Copper(I) chloride is the reductant, and oxygen is the oxidant. b. 46 pounds c. temperature, humidity, and wind (to bring more O2 into contact with the statue) 27. 2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g) Sulfur in PbS has been oxidized, and oxygen has been reduced. 1.1 × 103 g SO2 is produced. Lead is a toxic metal. Sulfur dioxide reacts with water to make acid rain. 29. 10.8 g benzene; limiting reactant is benzoic acid; 7.9 g of benzene 31. a. SiO2 + 6HF → SiF62− + 2H+ + 2H2O b. 2.8 g c. 6400 g HF 33. a. b. c. d.
C8H3 C8H3 C4H2; 6.0 g Complex molecules are essential for life. Reactions that help block UV may have implications regarding life on other planets.
35. The disaster occurred because organic compounds are highly flammable in a pure oxygen atmosphere. Using a mixture of 20% O2 and an inert gas such as N2 or He would have prevented the disaster.
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Chapter 4 Reactions in Aqueous Solution In Chapter 3 "Chemical Reactions", we described chemical reactions in general and introduced some techniques that are used to characterize them quantitatively. For the sake of simplicity, we discussed situations in which the reactants and the products of a given reaction were the only chemical species present. In reality, however, virtually every chemical reaction that takes place within and around us, such as the oxidation of foods to generate energy or the treatment of an upset stomach with an antacid tablet, occur in solution. In fact, many reactions must be carried out in solution and do not take place at all if the solid reactants are simply mixed.
The reaction of mercury(II) acetate with sodium iodide. When colorless aqueous solutions of each reactant are mixed, they produce a red precipitate, mercury(II) iodide, which is the result of an exchange reaction.
1. A homogeneous mixture of two or more substances in which solutes are dispersed uniformly throughout the solvent.
As you learned in Chapter 1 "Introduction to Chemistry", a solution1 is a homogeneous mixture in which substances present in lesser amounts, called
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solutes2, are dispersed uniformly throughout the substance in the greater amount, the solvent3. An aqueous solution4 is a solution in which the solvent is water, whereas in a nonaqueous solution, the solvent is a substance other than water. Familiar examples of nonaqueous solvents are ethyl acetate, used in nail polish removers, and turpentine, used to clean paint brushes. In this chapter, we focus on reactions that occur in aqueous solution. There are many reasons for carrying out reactions in solution. For a chemical reaction to occur, individual atoms, molecules, or ions must collide, and collisions between two solids, which are not dispersed at the atomic, molecular, or ionic level, do not occur at a significant rate. In addition, when the amount of a substance required for a reaction is so small that it cannot be weighed accurately, using a solution of that substance, in which the solute is dispersed in a much larger mass of solvent, enables chemists to measure its quantity with great precision. Chemists can also more effectively control the amount of heat consumed or produced in a reaction when the reaction occurs in solution, and sometimes the nature of the reaction itself can be controlled by the choice of solvent. This chapter introduces techniques for preparing and analyzing aqueous solutions, for balancing equations that describe reactions in solution, and for solving problems using solution stoichiometry. By the time you complete this chapter, you will know enough about aqueous solutions to explain what causes acid rain, why acid rain is harmful, and how a Breathalyzer measures alcohol levels. You will also understand the chemistry of photographic development, be able to explain why rhubarb leaves are toxic, and learn about a possible chemical reason for the decline and fall of the Roman Empire.
2. The substance or substances present in lesser amounts in a solution. 3. The substance present in the greater amount in a solution. 4. A solution in which water is the solvent.
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4.1 Aqueous Solutions LEARNING OBJECTIVE 1. To understand how and why solutions form.
The solvent in aqueous solutions is water, which makes up about 70% of the mass of the human body and is essential for life. Many of the chemical reactions that keep us alive depend on the interaction of water molecules with dissolved compounds. Moreover, as we will discuss in Chapter 5 "Energy Changes in Chemical Reactions", the presence of large amounts of water on Earth’s surface helps maintain its surface temperature in a range suitable for life. In this section, we describe some of the interactions of water with various substances and introduce you to the characteristics of aqueous solutions.
Polar Substances As shown in Figure 4.1 "The Polar Nature of Water", the individual water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. As is typical of group 16 elements, the oxygen atom in each O–H covalent bond attracts electrons more strongly than the hydrogen atom does. (For more information on periodic table groups and covalent bonding, see Chapter 2 "Molecules, Ions, and Chemical Formulas" and Chapter 7 "The Periodic Table and Periodic Trends".) Consequently, the oxygen and hydrogen nuclei do not equally share electrons. Instead, hydrogen atoms are electron poor compared with a neutral hydrogen atom and have a partial positive charge, which is indicated by δ +. The oxygen atom, in contrast, is more electron rich than a neutral oxygen atom, so it has a partial negative charge. This charge must be twice as large as the partial positive charge on each hydrogen for the molecule to have a net charge of zero. Thus its charge is indicated by 2δ−. This unequal distribution of charge creates a polar bond5, in which one portion of the molecule carries a partial negative charge, while the other portion carries a partial positive charge (Figure 4.1 "The Polar Nature of Water"). Because of the arrangement of polar bonds in a water molecule, water is described as a polar substance.
5. A chemical bond in which there is an unequal distribution of charge between the bonding atoms.
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Figure 4.1 The Polar Nature of Water
Each water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. Because the oxygen atom attracts electrons more strongly than the hydrogen atoms do, the oxygen atom is partially negatively charged (2δ−; blue) and the hydrogen atoms are partially positively charged (δ +; red). For the molecule to have a net charge of zero, the partial negative charge on oxygen must be twice as large as the partial positive charge on each hydrogen.
Because of the asymmetric charge distribution in the water molecule, adjacent water molecules are held together by attractive electrostatic (δ+…δ−) interactions between the partially negatively charged oxygen atom of one molecule and the partially positively charged hydrogen atoms of adjacent molecules (Figure 4.2 "The Structure of Liquid Water"). Energy is needed to overcome these electrostatic attractions. In fact, without them, water would evaporate at a much lower temperature, and neither Earth’s oceans nor we would exist!
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Figure 4.2 The Structure of Liquid Water
Two views of a water molecule are shown: (a) a ball-and-stick structure and (b) a space-filling model. Water molecules are held together by electrostatic attractions (dotted lines) between the partially negatively charged oxygen atom of one molecule and the partially positively charged hydrogen atoms on adjacent molecules. As a result, the water molecules in liquid water form transient networks with structures similar to that shown. Because the interactions between water molecules are continually breaking and reforming, liquid water does not have a single fixed structure.
As you learned in Section 2.1 "Chemical Compounds", ionic compounds such as sodium chloride (NaCl) are also held together by electrostatic interactions—in this case, between oppositely charged ions in the highly ordered solid, where each ion is surrounded by ions of the opposite charge in a fixed arrangement. In contrast to an ionic solid, the structure of liquid water is not completely ordered because the interactions between molecules in a liquid are constantly breaking and reforming. The unequal charge distribution in polar liquids such as water makes them good solvents for ionic compounds. When an ionic solid dissolves in water, the ions dissociate. That is, the partially negatively charged oxygen atoms of the H 2O molecules surround the cations (Na+ in the case of NaCl), and the partially positively charged hydrogen atoms in H2O surround the anions (Cl−; Figure 4.3 "The Dissolution of Sodium Chloride in Water"). Individual cations and anions that are each surrounded by their own shell of water molecules are called hydrated ions6. We can describe the dissolution of NaCl in water as Equation 4.1 6. Individual cations and anions that are each surrounded by their own shell of water molecules.
4.1 Aqueous Solutions
H O(l)
2 NaCl(s) ⎯⎯⎯⎯⎯ → Na+ (aq) + Cl– (aq)
where (aq) indicates that Na+ and Cl− are hydrated ions.
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Figure 4.3 The Dissolution of Sodium Chloride in Water
An ionic solid such as sodium chloride dissolves in water because of the electrostatic attraction between the cations (Na+) and the partially negatively charged oxygen atoms of water molecules, and between the anions (Cl −) and the partially positively charged hydrogen atoms of water.
Note the Pattern Polar liquids are good solvents for ionic compounds.
Electrolytes When electricity, in the form of an electrical potential, is applied to a solution, ions in solution migrate toward the oppositely charged rod or plate to complete an electrical circuit, whereas neutral molecules in solution do not (Figure 4.4 "The Effect of Ions on the Electrical Conductivity of Water"). Thus solutions that contain ions conduct electricity, while solutions that contain only neutral molecules do not.
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Electrical current will flow through the circuit shown in Figure 4.4 "The Effect of Ions on the Electrical Conductivity of Water" and the bulb will glow only if ions are present. The lower the concentration of ions in solution, the weaker the current and the dimmer the glow. Pure water, for example, contains only very low concentrations of ions, so it is a poor electrical conductor.
Note the Pattern Solutions that contain ions conduct electricity.
Figure 4.4 The Effect of Ions on the Electrical Conductivity of Water
An electrical current will flow and light the bulb only if the solution contains ions. (a) Pure water or an aqueous solution of a nonelectrolyte allows almost no current to flow, and the bulb does not light. (b) A weak electrolyte produces a few ions, allowing some current to flow and the bulb to glow dimly. (c) A strong electrolyte produces many ions, allowing more current to flow and the bulb to shine brightly.
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An electrolyte7 is any compound that can form ions when it dissolves in water. When strong electrolytes8 dissolve, the constituent ions dissociate completely due to strong electrostatic interactions with the solvent, producing aqueous solutions that conduct electricity very well (Figure 4.4 "The Effect of Ions on the Electrical Conductivity of Water"). Examples include ionic compounds such as barium chloride (BaCl2) and sodium hydroxide (NaOH), which are both strong electrolytes and dissociate as follows: Equation 4.2 H O(l)
2 BaCl2 (s) ⎯⎯⎯⎯⎯ → Ba2+ (aq) + 2Cl – (aq)
Equation 4.3 H O(l)
2 NaOH(s) ⎯⎯⎯⎯⎯ → Na+ (aq) + OH– (aq)
The single arrows from reactant to products in Equation 4.2 and Equation 4.3 indicate that dissociation is complete.
7. Any compound that can form ions when dissolved in water (c.f. nonelectrolytes). Electrolytes may be strong or weak. 8. An electrolyte that dissociates completely into ions when dissolved in water, thus producing an aqueous solution that conducts electricity very well. 9. A compound that produces relatively few ions when dissolved in water, thus producing an aqueous solution that conducts electricity poorly. 10. A substance that dissolves in water to form neutral molecules and has essentially no effect on electrical conductivity.
4.1 Aqueous Solutions
When weak electrolytes9 dissolve, they produce relatively few ions in solution. This does not mean that the compounds do not dissolve readily in water; many weak electrolytes contain polar bonds and are therefore very soluble in a polar solvent such as water. They do not completely dissociate to form ions, however, because of their weaker electrostatic interactions with the solvent. Because very few of the dissolved particles are ions, aqueous solutions of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes. One such compound is acetic acid (CH3CO2H), which contains the –CO2H unit. Although it is soluble in water, it is a weak acid and therefore also a weak electrolyte. Similarly, ammonia (NH 3) is a weak base and therefore a weak electrolyte. The behavior of weak acids and weak bases will be described in more detail when we discuss acid–base reactions in Section 4.6 "Acid–Base Reactions". Nonelectrolytes10 that dissolve in water do so as neutral molecules and thus have essentially no effect on conductivity. Examples of nonelectrolytes that are very soluble in water but that are essentially nonconductive are ethanol, ethylene glycol, glucose, and sucrose, all of which contain the –OH group that is characteristic of alcohols. In Chapter 8 "Ionic versus Covalent Bonding", we will discuss why alcohols and carboxylic acids behave differently in aqueous solution; for now, however, you can simply look for the presence of the –OH and –CO2H groups when trying to predict whether a substance is a strong electrolyte, a weak electrolyte, or a nonelectrolyte. In addition to alcohols, two other classes of organic compounds that
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are nonelectrolytes are aldehydes11 and ketones12, whose general structures are shown here. The distinctions between soluble and insoluble substances and between strong, weak, and nonelectrolytes are illustrated in Figure 4.5 "The Difference between Soluble and Insoluble Compounds (a) and Strong, Weak, and Nonelectrolytes (b)".
Note the Pattern Ionic substances and carboxylic acids are electrolytes; alcohols, aldehydes, and ketones are nonelectrolytes.
11. A class of organic compounds that has the general form RCHO, in which the carbon atom of the carbonyl group is bonded to a hydrogen atom and an R group. The R group may be either another hydrogen atom or an alkyl group (c.f. ketone). 12. A class of organic compounds with the general form RC(O)R’, in which the carbon atom of the carbonyl group is bonded to two alkyl groups (c.f. aldehyde). The alkyl groups may be the same or different.
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Figure 4.5 The Difference between Soluble and Insoluble Compounds (a) and Strong, Weak, and Nonelectrolytes (b)
General structure of an aldehyde and a ketone. Notice that both contain the C=O group.
When a soluble compound dissolves, its constituent atoms, molecules, or ions disperse throughout the solvent. In contrast, the constituents of an insoluble compound remain associated with one another in the solid. A soluble compound is a strong electrolyte if it dissociates completely into ions, a weak electrolyte if it dissociates only slightly into ions, and a nonelectrolyte if it dissolves to produce only neutral molecules.
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EXAMPLE 1 Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.
a. formaldehyde
b. cesium chloride Given: compound Asked for: relative ability to form ions in water Strategy: A Classify the compound as ionic or covalent. B If the compound is ionic and dissolves, it is a strong electrolyte that will dissociate in water completely to produce a solution that conducts electricity well. If the compound is covalent and organic, determine whether it contains the carboxylic acid group. If the compound contains this group, it is a weak electrolyte. If not, it is a nonelectrolyte. Solution: a. A Formaldehyde is an organic compound, so it is covalent. B It contains an aldehyde group, not a carboxylic acid group, so it should be a nonelectrolyte.
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b. A Cesium chloride (CsCl) is an ionic compound that consists of Cs + and Cl− ions. B Like virtually all other ionic compounds that are soluble in water, cesium chloride will dissociate completely into Cs +(aq) and Cl−(aq) ions. Hence it should be a strong electrolyte. Exercise Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.
a. (CH3)2CHOH (2-propanol)
b. ammonium sulfate Answer: a. nonelectrolyte b. strong electrolyte
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Summary Most chemical reactions are carried out in solutions, which are homogeneous mixtures of two or more substances. In a solution, a solute (the substance present in the lesser amount) is dispersed in a solvent (the substance present in the greater amount). Aqueous solutions contain water as the solvent, whereas nonaqueous solutions have solvents other than water. Polar substances, such as water, contain asymmetric arrangements of polar bonds, in which electrons are shared unequally between bonded atoms. Polar substances and ionic compounds tend to be most soluble in water because they interact favorably with its structure. In aqueous solution, dissolved ions become hydrated; that is, a shell of water molecules surrounds them. Substances that dissolve in water can be categorized according to whether the resulting aqueous solutions conduct electricity. Strong electrolytes dissociate completely into ions to produce solutions that conduct electricity well. Weak electrolytes produce a relatively small number of ions, resulting in solutions that conduct electricity poorly. Nonelectrolytes dissolve as uncharged molecules and have no effect on the electrical conductivity of water.
KEY TAKEAWAY • Aqueous solutions can be classified as polar or nonpolar depending on how well they conduct electricity.
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CONCEPTUAL PROBLEMS 1. What are the advantages to carrying out a reaction in solution rather than simply mixing the pure reactants? 2. What types of compounds dissolve in polar solvents? 3. Describe the charge distribution in liquid water. How does this distribution affect its physical properties? 4. Must a molecule have an asymmetric charge distribution to be polar? Explain your answer. 5. Why are many ionic substances soluble in water? 6. Explain the phrase like dissolves like. 7. What kinds of covalent compounds are soluble in water? 8. Why do most aromatic hydrocarbons have only limited solubility in water? Would you expect their solubility to be higher, lower, or the same in ethanol compared with water? Why? 9. Predict whether each compound will dissolve in water and explain why. a. b. c. d. e.
toluene acetic acid sodium acetate butanol pentanoic acid
10. Predict whether each compound will dissolve in water and explain why. a. b. c. d. e.
ammonium chloride 2-propanol heptane potassium dichromate 2-octanol
11. Given water and toluene, predict which is the better solvent for each compound and explain your reasoning. a. b. c. d.
4.1 Aqueous Solutions
sodium cyanide benzene acetic acid sodium ethoxide (CH3CH2ONa)
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12. Of water and toluene, predict which is the better solvent for each compound and explain your reasoning. a. b. c. d.
t-butanol calcium chloride sucrose cyclohexene
13. Compound A is divided into three equal samples. The first sample does not dissolve in water, the second sample dissolves only slightly in ethanol, and the third sample dissolves completely in toluene. What does this suggest about the polarity of A? 14. You are given a mixture of three solid compounds—A, B, and C—and are told that A is a polar compound, B is slightly polar, and C is nonpolar. Suggest a method for separating these three compounds. 15. A laboratory technician is given a sample that contains only sodium chloride, sucrose, and cyclodecanone (a ketone). You must tell the technician how to separate these three compounds from the mixture. What would you suggest? 16. Many over-the-counter drugs are sold as ethanol/water solutions rather than as purely aqueous solutions. Give a plausible reason for this practice. 17. What distinguishes a weak electrolyte from a strong electrolyte? 18. Which organic groups result in aqueous solutions that conduct electricity? 19. It is considered highly dangerous to splash barefoot in puddles during a lightning storm. Why? 20. Which solution(s) would you expect to conduct electricity well? Explain your reasoning. a. b. c. d.
an aqueous solution of sodium chloride a solution of ethanol in water a solution of calcium chloride in water a solution of sucrose in water
21. Which solution(s) would you expect to conduct electricity well? Explain your reasoning. a. b. c. d.
4.1 Aqueous Solutions
an aqueous solution of acetic acid an aqueous solution of potassium hydroxide a solution of ethylene glycol in water a solution of ammonium chloride in water
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22. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning. a. b. c. d.
potassium hydroxide ammonia calcium chloride butanoic acid
23. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning. a. b. c. d.
magnesium hydroxide butanol ammonium bromide pentanoic acid
24. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in aqueous solution? Explain your reasoning. a. b. c. d. e.
4.1 Aqueous Solutions
H2SO4 diethylamine 2-propanol ammonium chloride propanoic acid
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ANSWERS 5. Ionic compounds such as NaCl are held together by electrostatic interactions between oppositely charged ions in the highly ordered solid. When an ionic compound dissolves in water, the partially negatively charged oxygen atoms of the H2O molecules surround the cations, and the partially positively charged hydrogen atoms in H2O surround the anions. The favorable electrostatic interactions between water and the ions compensate for the loss of the electrostatic interactions between ions in the solid. 9. a. Because toluene is an aromatic hydrocarbon that lacks polar groups, it is unlikely to form a homogenous solution in water. b. Acetic acid contains a carboxylic acid group attached to a small alkyl group (a methyl group). Consequently, the polar characteristics of the carboxylic acid group will be dominant, and acetic acid will form a homogenous solution with water. c. Because most sodium salts are soluble, sodium acetate should form a homogenous solution with water. d. Like all alcohols, butanol contains an −OH group that can interact well with water. The alkyl group is rather large, consisting of a 4-carbon chain. In this case, the nonpolar character of the alkyl group is likely to be as important as the polar character of the –OH, decreasing the likelihood that butanol will form a homogeneous solution with water. e. Like acetic acid, pentanoic acid is a carboxylic acid. Unlike acetic acid, however, the alkyl group is rather large, consisting of a 4-carbon chain as in butanol. As with butanol, the nonpolar character of the alkyl group is likely to be as important as the polar character of the carboxylic acid group, making it unlikely that pentanoic acid will form a homogeneous solution with water. (In fact, the solubility of both butanol and pentanoic acid in water is quite low, only about 3 g per 100 g water at 25°C.) 17. An electrolyte is any compound that can form ions when it dissolves in water. When a strong electrolyte dissolves in water, it dissociates completely to give the constituent ions. In contrast, when a weak electrolyte dissolves in water, it produces relatively few ions in solution.
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4.2 Solution Concentrations LEARNING OBJECTIVE 1. To describe the concentrations of solutions quantitatively.
All of us have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration13 of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution. Chemists use many different ways to define concentrations, some of which are described in this section.
Molarity The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M)14 of a solution is the number of moles of solute present in exactly 1 L of solution. Molarity is also the number of millimoles of solute present in exactly 1 mL of solution: Equation 4.4
molarity =
13. The quantity of solute that is dissolved in a particular quantity of solvent or solution. 14. A common unit of concentration that is the number of moles of solute present in exactly 1 L of solution (mol/L) .
moles of solute mmoles of solute = liters of solution milliliters of solution
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as M. An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So [sucrose] = 1.00 M is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
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Equation 4.5
V L Mmol/L = L
( L ) mol
= moles
or Equation 4.6
V mL Mmmol/mL = mL
( mL ) mmol
= mmoles
Example 2 illustrates the use of Equation 4.5 and Equation 4.6.
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EXAMPLE 2 Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH. Given: identity of solute and volume and molarity of solution Asked for: amount of solute in moles Strategy: Use either Equation 4.5 or Equation 4.6, depending on the units given in the problem. Solution: Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.5 is more useful:
moles NaOH = V L Mmol/L = (2.50 L )
(
0.100 mol L
)
= 0.250 mol NaO
Exercise Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine. Answer: 41.6 mmol
Concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-tovolume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per
4.2 Solution Concentrations
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milliliter. These concentrations and their units are summarized in Table 4.1 "Common Units of Concentration". Table 4.1 Common Units of Concentration Concentration
Units
m/m
g of solute/g of solution
m/v
g of solute/mL of solution
ppm
ppb
g of solute/106 g of solution μg/mL g of solute/109 g of solution ng/mL
The Preparation of Solutions To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute" illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure 4.7 "Preparation of 250 mL of a Solution of (NH", for some substances this effect can be significant, especially for concentrated solutions. Figure 4.6 Preparation of a Solution of Known Concentration Using a Solid Solute
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Figure 4.7 Preparation of 250 mL of a Solution of (NH4)2Cr2O7 in Water
The solute occupies space in the solution, so less than 250 mL of water are needed to make 250 mL of solution.
4.2 Solution Concentrations
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EXAMPLE 3 The solution in Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute" contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2·2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl2·2H2O? Given: mass of solute and volume of solution Asked for: concentration (M) Strategy: To find the number of moles of CoCl2·2H2O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters. Solution: The molar mass of CoCl2·2H2O is 165.87 g/mol. Therefore,
moles CoCl 2 •2H 2 O =
10.0 g
( 165.87 g /mol )
= 0.0603 mol
The volume of the solution in liters is
volume = 500 mL
( 1000 mL ) 1L
= 0.500 L
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
molarity =
0.0603 mol = 0.121 M = CoCl 2 •H2 O 0.500 L
Exercise
4.2 Solution Concentrations
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The solution shown in Figure 4.7 "Preparation of 250 mL of a Solution of (NH" contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate? Answer: (NH4)2Cr2O7 = 1.43 M
To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation 4.5. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 4.
4.2 Solution Concentrations
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EXAMPLE 4 The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol. Given: molarity, volume, and molar mass of solute Asked for: mass of solute Strategy: A Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity. B Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass. Solution: A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:
V L Mmol/L = moles 500 mL
1 L
( 1000 mL ) (
0.310 mol glucose 1L
)
= 0.155 mol glucose
B We then convert the number of moles of glucose to the required mass of glucose:
180.16 g glucose mass of glucose = 0.155 mol glucose 1 mol glucose
= 27.9 g gl
Exercise
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Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution. Answer: 2.3 g NaCl
A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution15, which is a commercially prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids. The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 4.8 "Preparation of a Solution of Known Concentration by Diluting a Stock Solution". It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore Equation 4.7 (Vs)(Ms) = moles of solute = (Vd)(Md) where the subscripts s and d indicate the stock and dilute solutions, respectively. Example 5 demonstrates the calculations involved in diluting a concentrated stock solution.
15. A commercially prepared solution of known concentration.
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Figure 4.8 Preparation of a Solution of Known Concentration by Diluting a Stock Solution
(a) A volume (Vs) containing the desired moles of solute (Ms) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [(Vs)(Ms) = (Vd)(Md)].
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EXAMPLE 5 What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4? Given: volume and molarity of dilute solution Asked for: volume of stock solution Strategy: A Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity. B To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution. Solution: A The D5W solution in Example 4 was 0.310 M glucose. We begin by using Equation 4.7 to calculate the number of moles of glucose contained in 2500 mL of the solution:
moles glucose = 2500 mL
1 L
( 1000 mL ) (
)
0.310 mol glucose 1 L
B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:
1L volume of stock soln = 0.775 mol glucose 3.00 mol glucose
= 0.2
In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed.
4.2 Solution Concentrations
= 0
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Chapter 4 Reactions in Aqueous Solution
The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M). We could also have solved this problem in a single step by solving Equation 4.7 for Vs and substituting the appropriate values:
Vs =
(2.500 L)(0.310 M ) (V d )(M d ) = = 0.258 L Ms 3.00 M
As we have noted, there is often more than one correct way to solve a problem. Exercise What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)? Answer: 16 mL
Ion Concentrations in Solution In Example 3, you calculated that the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL is 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH4+ ions and one Cr2O72− ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72− ions: Equation 4.8 H O(l)
–
2 (NH4 )2 Cr2 O7 (s) ⎯⎯⎯⎯⎯ → 2NH4 + (aq) + Cr2 O7 2 (aq)
Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72− anions and 2 mol of NH4+ cations (see Figure 4.9 "Dissolution of 1 mol of an Ionic Compound").
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Figure 4.9 Dissolution of 1 mol of an Ionic Compound
In this case, dissolving 1 mol of (NH4)2Cr2O7 produces a solution that contains 1 mol of Cr2O72− ions and 2 mol of NH4+ ions. (Water molecules are omitted from a molecular view of the solution for clarity.)
When we carry out a chemical reaction using a solution of a salt such as ammonium dichromate, we need to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH4)2Cr2O7, then the concentration of Cr2O72− must also be 1.43 M because there is one Cr2O72− ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH4)2Cr2O7 produces three ions when dissolved in water (2NH4+ + 1Cr2O72−), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.
4.2 Solution Concentrations
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EXAMPLE 6 What are the concentrations of all species derived from the solutes in these aqueous solutions? a. 0.21 M NaOH b. 3.7 M (CH3)CHOH c. 0.032 M In(NO3)3 Given: molarity Asked for: concentrations Strategy: A Classify each compound as either a strong electrolyte or a nonelectrolyte. B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution. Solution:
a. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution: H O(l)
2 NaOH(s) ⎯⎯⎯⎯⎯ → Na + (aq) + OH – (aq)
B Because each formula unit of NaOH produces one Na+ ion and one OH− ion, the concentration of each ion is the same as the concentration of NaOH: [Na+] = 0.21 M and [OH−] = 0.21 M. b. A The formula (CH3)2CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 "Aqueous Solutions" that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.
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B The only solute species in solution is therefore (CH3)2CHOH molecules, so [(CH3)2CHOH] = 3.7 M. c. A Indium nitrate is an ionic compound that contains In3+ ions and NO3− ions, so we expect it to behave like a strong electrolyte in aqueous solution: H O(l)
2 In(NO 3 )3 (s) ⎯⎯⎯⎯⎯ → In 3+ (aq) + 3NO 3 – (aq)
B One formula unit of In(NO3)3 produces one In3+ ion and three NO3− ions, so a 0.032 M In(NO3)3 solution contains 0.032 M In3+ and 3 × 0.032 M = 0.096 M NO3–—that is, [In3+] = 0.032 M and [NO3−] = 0.096 M. Exercise What are the concentrations of all species derived from the solutes in these aqueous solutions? a. 0.0012 M Ba(OH)2 b. 0.17 M Na2SO4 c. 0.50 M (CH3)2CO, commonly known as acetone
Answer: a. [Ba2+] = 0.0012 M; [OH−] = 0.0024 M b. [Na+] = 0.34 M; [SO42−] = 0.17 M c. [(CH3)2CO] = 0.50 M
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KEY EQUATIONS definition of molarity Equation 4.4: molarity
=
moles of solute liters of solution
=
mmoles of solute milliliters of solution
relationship among volume, molarity, and moles Equation 4.5: V L M mol/L
= L
mol ( L
)
= moles
relationship between volume and concentration of stock and dilute solutions Equation 4.7: (Vs)(Ms) = moles of solute = (Vd)(Md)
Summary The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed as molarity, the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.
KEY TAKEAWAY • Solution concentrations are typically expressed as molarity and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.
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CONCEPTUAL PROBLEMS 1. Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers. a. NH3 b. HF c. CH3CH2CH2OH d. Na2SO4
2. Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers. a. b. c. d. e.
CH3CO2H NaCl Na2S Na3PO4 acetaldehyde
3. Would you expect a 1.0 M solution of CaCl2 to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not? 4. An alternative way to define the concentration of a solution is molality, abbreviated m. Molality is defined as the number of moles of solute in 1 kg of solvent. How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 m solution of sucrose? Explain your answer. 5. What are the advantages of using solutions for quantitative calculations?
ANSWER 5. If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately.
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NUMERICAL PROBLEMS 1. Calculate the number of grams of solute in 1.000 L of each solution. a. b. c. d.
0.2593 M NaBrO3 1.592 M KNO3 1.559 M acetic acid 0.943 M potassium iodate
2. Calculate the number of grams of solute in 1.000 L of each solution. a. b. c. d.
0.1065 M BaI2 1.135 M Na2SO4 1.428 M NH4Br 0.889 M sodium acetate
3. If all solutions contain the same solute, which solution contains the greater mass of solute? a. 1.40 L of a 0.334 M solution or 1.10 L of a 0.420 M solution b. 25.0 mL of a 0.134 M solution or 10.0 mL of a 0.295 M solution c. 250 mL of a 0.489 M solution or 150 mL of a 0.769 M solution 4. Complete the following table for 500 mL of solution. Compound calcium sulfate
Mass (g)
Moles
Concentration (M)
4.86
acetic acid
3.62
hydrogen iodide dihydrate barium bromide
1.273 3.92
glucose sodium acetate
0.983 2.42
5. What is the concentration of each species present in the following aqueous solutions? a. b. c. d.
4.2 Solution Concentrations
0.489 mol of NiSO4 in 600 mL of solution 1.045 mol of magnesium bromide in 500 mL of solution 0.146 mol of glucose in 800 mL of solution 0.479 mol of CeCl3 in 700 mL of solution
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6. What is the concentration of each species present in the following aqueous solutions? a. b. c. d.
0.324 mol of K2MoO4 in 250 mL of solution 0.528 mol of potassium formate in 300 mL of solution 0.477 mol of KClO3 in 900 mL of solution 0.378 mol of potassium iodide in 750 mL of solution
7. What is the molar concentration of each solution? a. b. c. d.
8.7 g of calcium bromide in 250 mL of solution 9.8 g of lithium sulfate in 300 mL of solution 12.4 g of sucrose (C12H22O11) in 750 mL of solution 14.2 g of iron(III) nitrate hexahydrate in 300 mL of solution
8. What is the molar concentration of each solution? a. b. c. d.
12.8 g of sodium hydrogen sulfate in 400 mL of solution 7.5 g of potassium hydrogen phosphate in 250 mL of solution 11.4 g of barium chloride in 350 mL of solution 4.3 g of tartaric acid (C4H6O6) in 250 mL of solution
9. Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant. a. b. c. d. e.
BaCl2(aq) + Na2SO4(aq) → Ca(OH)2(aq) + H3PO4(aq) → Al(NO3)3(aq) + H2SO4(aq) → Pb(NO3)2(aq) + CuSO4(aq) → Al(CH3CO2)3(aq) + NaOH(aq) →
10. An experiment required 200.0 mL of a 0.330 M solution of Na 2CrO4. A stock solution of Na2CrO4 containing 20.0% solute by mass with a density of 1.19 g/ cm3 was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na2CrO4 using the stock solution. 11. Calcium hypochlorite [Ca(OCl)2] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl)2 concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite? 12. Phenol (C6H5OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol?
4.2 Solution Concentrations
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Chapter 4 Reactions in Aqueous Solution
13. If a tablet containing 100 mg of caffeine (C8H10N4O2) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution? 14. A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered?
ANSWERS 11. 0.48 M ClO− 13. 1.74 × 10−3 M caffeine
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4.3 Stoichiometry of Reactions in Solution LEARNING OBJECTIVES 1. To balance equations that describe reactions in solution. 2. To solve quantitative problems involving the stoichiometry of reactions in solution.
Quantitative calculations involving reactions in solution are carried out in the same manner as we discussed in Chapter 3 "Chemical Reactions". Instead of masses, however, we use volumes of solutions of known concentration to determine the number of moles of reactants. Whether we are dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation tell us the number of moles of each reactant needed and the number of moles of each product that can be produced.
Calculating Moles from Volume An expanded version of the flowchart for stoichiometric calculations illustrated in Figure 3.5 "Steps for Obtaining an Empirical Formula from Combustion Analysis" is shown in Figure 4.10 "An Expanded Flowchart for Stoichiometric Calculations". We can use the balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products to determine the amounts of other species, as illustrated in Example 7, Example 8, and Example 9. Figure 4.10 An Expanded Flowchart for Stoichiometric Calculations
Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in a balanced chemical equation.
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Note the Pattern The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.
4.3 Stoichiometry of Reactions in Solution
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EXAMPLE 7 Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2]− ion. Gold is then recovered by reduction with metallic zinc according to the following equation: Zn(s) + 2[Au(CN)2]−(aq) → [Zn(CN)4]2−(aq) + 2Au(s) What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10−4 M solution of [Au(CN)2]−? Given: chemical equation and molarity and volume of reactant Asked for: mass of product Strategy: A Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN) 2]− present by multiplying the volume of the solution by its concentration. B From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass. Solution: A The equation is balanced as written, so we can proceed to the stoichiometric calculation. We can adapt Figure 4.10 "An Expanded Flowchart for Stoichiometric Calculations" for this particular problem as follows:
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Chapter 4 Reactions in Aqueous Solution
As indicated in the strategy, we start by calculating the number of moles of [Au(CN)2]− present in the solution from the volume and concentration of the [Au(CN)2]− solution:
moles [Au(CN) 2 ]–
= V L Mmol/L = 400.0 L
(
3.30 × 10 4– mol [Au(CN) 2 ]– 1 L
)
B Because the coefficients of gold and the [Au(CN)2]− ion are the same in the balanced chemical equation, if we assume that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)2]− we started with (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so we need to convert the number of moles of gold to the corresponding mass using the molar mass of gold:
mass of Au
= (moles Au)(molar mass Au) = 0.132 mol Au
( 1 mol Au ) 196.97 g Au
= 26.0 g Au
At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.
26.0 g Au ×
1 troy oz 31.10 g
×
$1400 1 troy oz Au
= $1170
Exercise What mass of solid lanthanum(III) oxalate nonahydrate [La 2(C2O4)3·9H2O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl 3 by adding a stoichiometric amount of sodium oxalate? Answer: 3.89 g
Limiting Reactants in Solutions The concept of limiting reactants applies to reactions that are carried out in solution as well as to reactions that involve pure substances. If all the reactants but
4.3 Stoichiometry of Reactions in Solution
442
= 0
Chapter 4 Reactions in Aqueous Solution
one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example 8.
4.3 Stoichiometry of Reactions in Solution
443
Chapter 4 Reactions in Aqueous Solution
EXAMPLE 8 Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion: Ag +
3CH 3 CH2 OH(aq) + 2Cr 2 O7 2– (aq) + 16H + (aq) ⎯H⎯⎯⎯⎯⎯⎯⎯⎯⎯ → 3CH 3 CO2 H(aq SO (aq) yellow-orange
2
4
When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr2O72− ions are reduced from Cr6+ to Cr3+. In the presence of Ag+ ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr2O72− ion (the reactant) is yellow-orange and the Cr3+ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.
A Breathalyzer ampul before (a) and after (b) ethanol is added. When a measured volume of a suspect’s breath is bubbled through the
4.3 Stoichiometry of Reactions in Solution
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Chapter 4 Reactions in Aqueous Solution
solution, the ethanol is oxidized to acetic acid, and the solution changes color from yellow-orange to green. The intensity of the green color indicates the amount of ethanol in the sample.
A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K2Cr2O7 in 50% H2SO4 as well as a fixed concentration of AgNO3 (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr 6+ to Cr3+? Given: volume and concentration of one reactant Asked for: mass of other reactant needed for complete reaction Strategy: A Calculate the number of moles of Cr2O72− ion in 1 mL of the Breathalyzer solution by dividing the mass of K2Cr2O7 by its molar mass. B Find the total number of moles of Cr2O72− ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL). C Use the mole ratios from the balanced chemical equation to calculate the number of moles of C2H5OH needed to react completely with the number of moles of Cr2O72− ions present. Then find the mass of C2H5OH needed by multiplying the number of moles of C2H5OH by its molar mass. Solution: A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K2Cr2O7 in 1 mL of solution, which we can use to calculate the number of moles of K2Cr2O7 contained in 1 mL:
4.3 Stoichiometry of Reactions in Solution
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Chapter 4 Reactions in Aqueous Solution
(0.25 mg K2 Cr2 O7 ) 1 g moles K2 Cr2 O7 1 mol = 1 mL mL ( 1000 mg ) ( 294.18 g K2 B Because 1 mol of K2Cr2O7 produces 1 mol of Cr2O72− when it dissolves, each milliliter of solution contains 8.5 × 10−7 mol of Cr2O72−. The total number of moles of Cr2O72− in a 3.0 mL Breathalyzer ampul is thus
moles Cr2 O7
2–
=
(
8.5 × 10 –7 mol 1 mL
)
(3.0 mL ) = 2.6 × 10 –6 mol Cr2 O
C The balanced chemical equation tells us that 3 mol of C 2H5OH is needed to consume 2 mol of Cr2O72− ion, so the total number of moles of C2H5OH required for complete reaction is
moles of C 2 H5 OH = (2.6 × 10
–6
mol Cr 2 O7
2–
3 mol C 2 H5 OH ) 2 mol Cr 2 O7 2–
As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass:
mass C2 H5 OH = (3.9 × 10
–6
46.07 g mol C2 H5 OH ) mol C2 H5 OH
= 1.8 ×
Thus 1.8 × 10−4 g or 0.18 mg of C2H5OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal. Exercise The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction
4.3 Stoichiometry of Reactions in Solution
=
446
Chapter 4 Reactions in Aqueous Solution
Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10−4 g of para-nitrophenol to ensure that formation of the yellow anion is complete? Answer: 4.93 × 10−5 L or 49.3 μL
In Example 7 and Example 8, the identity of the limiting reactant has been apparent: [Au(CN)2]−, LaCl3, ethanol, and para-nitrophenol. When the limiting reactant is not apparent, we can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation, just as we did in Chapter 3 "Chemical Reactions", Section 3.4 "Mass Relationships in Chemical Equations". The only difference is that now we use the volumes and concentrations of solutions of reactants rather than the masses of reactants to calculate the number of moles of reactants, as illustrated in Example 9.
4.3 Stoichiometry of Reactions in Solution
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Chapter 4 Reactions in Aqueous Solution
EXAMPLE 9 When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows: 2AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + 2KNO3(aq) What mass of Ag2Cr2O7 is formed when 500 mL of 0.17 M K2Cr2O7 are mixed with 250 mL of 0.57 M AgNO3? Given: balanced chemical equation and volume and concentration of each reactant Asked for: mass of product Strategy: A Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity. B Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation. C Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product. Solution: A The balanced chemical equation tells us that 2 mol of AgNO 3(aq) reacts with 1 mol of K2Cr2O7(aq) to form 1 mol of Ag2Cr2O7(s) (Figure 4.11 "What Happens at the Molecular Level When Solutions of AgNO"). The first step is to calculate the number of moles of each reactant in the specified volumes:
4.3 Stoichiometry of Reactions in Solution
448
Chapter 4 Reactions in Aqueous Solution
moles K2 Cr2 O7 = 500 mL moles AgNO 3 = 250 mL
1 L
( 1000 mL ) ( 1 L
( 1000 mL ) (
)
0.17 mol K 2 Cr2 O7 1 L
0.57 mol AgNO 3 1 L
)
= 0.1
B Now we can determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:
0.085 mol = 0.085 1 mol 0.14 mol AgNO 3 : = 0.070 2 mol K2 Cr2 O7 :
Because 0.070 < 0.085, we know that AgNO3 is the limiting reactant. C Each mole of Ag2Cr2O7 formed requires 2 mol of the limiting reactant (AgNO3), so we can obtain only 0.14/2 = 0.070 mol of Ag 2Cr2O7. Finally, we convert the number of moles of Ag2Cr2O7 to the corresponding mass:
mass of Ag2 Cr2 O7 = 0.070 mol
4.3 Stoichiometry of Reactions in Solution
( 1 mol ) 431.72 g
= 30 g Ag 2 Cr2 O7
449
=
Chapter 4 Reactions in Aqueous Solution
Figure 4.11 What Happens at the Molecular Level When Solutions of AgNO3 and K2Cr2O7 Are Mixed
The Ag+ and Cr2O72− ions form a red precipitate of solid Ag2Cr2O7, while the K+ and NO3− ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.)
Exercise Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation: 2NaHCO3(aq) + H2SO4(aq) → 2CO2(g) + Na2SO4(aq) + 2H2O(l) If 13.0 mL of 3.0 M H2SO4 are added to 732 mL of 0.112 M NaHCO3, what mass of CO2 is produced? Answer: 3.4 g
4.3 Stoichiometry of Reactions in Solution
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Chapter 4 Reactions in Aqueous Solution
Summary Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.
KEY TAKEAWAY • Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation.
CONCEPTUAL PROBLEMS 1. What information is required to determine the mass of solute in a solution if you know the molar concentration of the solution? 2. Is it possible for one reactant to be limiting in a reaction that does not go to completion?
4.3 Stoichiometry of Reactions in Solution
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Chapter 4 Reactions in Aqueous Solution
NUMERICAL PROBLEMS 1. Refer to the Breathalyzer test described in Example 8. How much ethanol must be present in 89.5 mL of a person’s breath to consume all the potassium dichromate in a Breathalyzer ampul containing 3.0 mL of a 0.40 mg/mL solution of potassium dichromate? 2. Phosphoric acid and magnesium hydroxide react to produce magnesium phosphate and water. If 45.00 mL of 1.50 M phosphoric acid are used in the reaction, how many grams of magnesium hydroxide are needed for the reaction to go to completion? 3. Barium chloride and sodium sulfate react to produce sodium chloride and barium sulfate. If 50.00 mL of 2.55 M barium chloride are used in the reaction, how many grams of sodium sulfate are needed for the reaction to go to completion? 4. How many grams of sodium phosphate are obtained in solution from the reaction of 75.00 mL of 2.80 M sodium carbonate with a stoichiometric amount of phosphoric acid? A second product is water; what is the third product? How many grams of the third product are obtained? 5. How many grams of ammonium bromide are produced from the reaction of 50.00 mL of 2.08 M iron(II) bromide with a stoichiometric amount of ammonium sulfide? What is the second product? How many grams of the second product are produced? 6. Lead(II) nitrate and hydroiodic acid react to produce lead(II) iodide and nitric acid. If 3.25 g of lead(II) iodide were obtained by adding excess HI to 150.0 mL of lead(II) nitrate, what was the molarity of the lead(II) nitrate solution? 7. Silver nitrate and sodium chloride react to produce sodium nitrate and silver chloride. If 2.60 g of AgCl was obtained by adding excess NaCl to 100 mL of AgNO3, what was the molarity of the silver nitrate solution?
4.3 Stoichiometry of Reactions in Solution
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4.4 Ionic Equations LEARNING OBJECTIVE 1. To understand what information is obtained by each type of ionic equation.
The chemical equations discussed in Chapter 3 "Chemical Reactions" showed the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent. Let’s consider the reaction of silver nitrate with potassium dichromate. As you learned in Example 9, when aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall chemical equation16 for the reaction shows each reactant and product as undissociated, electrically neutral compounds: Equation 4.9 2AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + 2KNO3(aq)
16. A chemical equation that shows all the reactants and products as undissociated, electrically neutral compounds. 17. A chemical equation that shows which ions and molecules are hydrated and which are present in other forms and phases.
Although Equation 4.9 gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO3 and K2Cr2O7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag2Cr2O7 is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation17, showing which ions and molecules are hydrated and which are present in other forms and phases: Equation 4.10 2Ag+(aq) + 2NO3−(aq) + 2K+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) + 2K+(aq) + 2NO3−(aq)
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Chapter 4 Reactions in Aqueous Solution
Note that K+(aq) and NO3−(aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ions18 because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation19, which shows only those species that participate in the chemical reaction: Equation 4.11 2Ag+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation 4.11, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag 2Cr2O7 formula unit. By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows: Equation 4.12 2AgF(aq) + (NH4)2Cr2O7(aq) → Ag2Cr2O7(s) + 2NH4F(aq) The complete ionic equation for this reaction is as follows: Equation 4.13 2Ag+(aq) + 2F−(aq) + 2NH4+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) + 2NH4+(aq) + 2F−(aq) Because two NH4+(aq) and two F−(aq) ions appear on both sides of Equation 4.13, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation 4.14), which is identical to Equation 4.11:
18. Ions that do not participate in the actual reaction. 19. A chemical equation that shows only those species that participate in the chemical reaction.
4.4 Ionic Equations
Equation 4.14 2Ag+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For
454
Chapter 4 Reactions in Aqueous Solution
example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.
4.4 Ionic Equations
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Chapter 4 Reactions in Aqueous Solution
EXAMPLE 10 Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate. Given: reactants and products Asked for: overall, complete ionic, and net ionic equations Strategy: Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation. Solution: From the information given, we can write the unbalanced chemical equation for the reaction: Ba(NO3)2(aq) + Na3PO4(aq) → Ba3(PO4)2(s) + NaNO3(aq) Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43− ions per formula unit, we can balance the equation by inspection: 3Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq) This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form: 3Ba2+(aq) + 6NO3−(aq) + 6Na+(aq) + 2PO43−(aq) → Ba3(PO4)2(s) + 6Na+(aq) + 6NO3−(aq) The six NO3−(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:
4.4 Ionic Equations
456
Chapter 4 Reactions in Aqueous Solution 3Ba2+(aq) + 2PO43−(aq) → Ba3(PO4)2(s) Exercise Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride. Answer: overall chemical equation: 3AgF(aq) + Na3PO4(aq) → Ag3PO4(s) + 3NaF(aq) complete ionic equation: 3Ag+(aq) + 3F−(aq) + 3Na+(aq) + PO43−(aq) → Ag3PO4(s) + 3Na+(aq) + 3F−(aq) net ionic equation: 3Ag+(aq) + PO43−(aq) → Ag3PO4(s)
So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome. The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions introduced in Chapter 3 "Chemical Reactions" (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.
4.4 Ionic Equations
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Summary The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.
KEY TAKEAWAY • A complete ionic equation consists of the net ionic equation and spectator ions.
CONCEPTUAL PROBLEM 1. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation?
4.4 Ionic Equations
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4.5 Precipitation Reactions LEARNING OBJECTIVE 1. To identify a precipitation reaction and predict solubilities.
A precipitation reaction20 is a reaction that yields an insoluble product—a precipitate21—when two solutions are mixed. In Section 4.4 "Ionic Equations", we described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate: Equation 4.15 AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + KNO3(aq) This equation has the general form of an exchange reaction: Equation 4.16
AC + BD → AD + BC insoluble
Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.
20. A subclass of an exchange reaction that yields an insoluble product (a precipitate) when two solutions are mixed.
Note the Pattern Precipitation reactions are a subclass of exchange reactions.
21. The insoluble product that forms in a precipitation reaction.
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Predicting Solubilities Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water" gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water" to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail in Chapter 17 "Solubility and Complexation Equilibriums", where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete. Table 4.2 Guidelines for Predicting the Solubility of Ionic Compounds in Water Soluble
Exceptions
most salts that contain an alkali Rule 1 metal (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+) Rule 2
most salts that contain the nitrate (NO3−) anion
most salts of anions derived Rule 3 from monocarboxylic acids (e.g., CH3CO2−) Rule 4
most chloride, bromide, and iodide salts Insoluble
but silver acetate and salts of long-chain not carboxylates salts of metal ions located on the lower but right side of the periodic table (e.g., Cu+, not Ag+, Pb2+, and Hg22+). Exceptions
salts of the alkali metals (group 1), the most salts that contain the but heavier alkaline earths (Ca2+, Sr2+, and Rule 5 hydroxide (OH−) and sulfide (S2−) not Ba2+ in group 2), and the NH4+ ion. anions Rule 6
most carbonate (CO32−) and phosphate (PO43−) salts
most sulfate (SO42−) salts that Rule 7 contain main group cations with a charge ≥ +2
4.5 Precipitation Reactions
but salts of the alkali metals or the NH4+ ion. not but salts of +1 cations, Mg2+, and dipositive not transition metal cations (e.g., Ni2+)
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Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na +(aq), 0.50 M Cl−(aq), 0.50 M K+(aq), and 0.50 M Br−(aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other (Figure 4.12 "The Effect of Mixing Aqueous KBr and NaCl Solutions"). Figure 4.12 The Effect of Mixing Aqueous KBr and NaCl Solutions
Because no net reaction occurs, the only effect is to dilute each solution with the other. (Water molecules are omitted from molecular views of the solutions for clarity.)
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EXAMPLE 11 Using the information in Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs. a. Aqueous solutions of barium chloride and lithium sulfate are mixed. b. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed. c. Aqueous solutions of strontium bromide and aluminum nitrate are mixed. d. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide. Given: reactants Asked for: reaction and net ionic equation Strategy: A Identify the ions present in solution and write the products of each possible exchange reaction. B Refer to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water" to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction. Solution:
a. A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba2+, Cl−, Li+, and SO42− ions. The only possible exchange reaction is to form LiCl and BaSO4:
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B We now need to decide whether either of these products is insoluble. Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water" shows that LiCl is soluble in water (rules 1 and 4), but BaSO4 is not soluble in water (rule 5). Thus BaSO4 will precipitate according to the net ionic equation Ba2+(aq) + SO42−(aq) → BaSO4(s) Although soluble barium salts are toxic, BaSO4 is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO4 particles in water.
An x-ray of the digestive organs of a patient who has swallowed a “barium milkshake.” A barium milkshake is a suspension of very fine BaSO4 particles in water; the high atomic mass of barium makes it opaque to x-rays.
b. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are
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mixed, the resulting solution initially contains Rb+, OH−, Co2+, and Cl− ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):
B According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", RbCl is soluble (rules 1 and 4), but Co(OH)2 is not soluble (rule 5). Hence Co(OH)2 will precipitate according to the following net ionic equation: Co2+(aq) + 2OH−(aq) → Co(OH)2(s) c. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr2+, Br−, Al3+, and NO3− ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:
B According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", both AlBr3 (rule 4) and Sr(NO3)2 (rule 2) are soluble. Thus no net reaction will occur. d. A According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb 2+ and CH3CO2− ions. Because the solution also contains NH4+ and I− ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:
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B According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", ammonium acetate is soluble (rules 1 and 3), but PbI2 is insoluble (rule 4). Thus Pb(C2H3O2)2 will dissolve, and PbI2 will precipitate. The net ionic equation is as follows: Pb2+ (aq) + 2I−(aq) → PbI2(s)
Exercise Using the information in Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs. a. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride. b. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate. c. Solid sodium fluoride is added to an aqueous solution of ammonium formate. d. Aqueous solutions of calcium bromide and cesium carbonate are mixed. Answer: a. b. c. d.
4.5 Precipitation Reactions
Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s) 2PO43−(aq) + 3Hg2+(aq) → Hg3(PO4)2(s) NaF(s) dissolves; no net reaction Ca2+(aq) + CO32−(aq) → CaCO3(s)
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Precipitation Reactions in Photography Precipitation reactions can be used to recover silver from solutions used to develop conventional photographic film. Although largely supplanted by digital photography, conventional methods are often used for artistic purposes. Silver bromide is an off-white solid that turns black when exposed to light, which is due to the formation of small particles of silver metal. Black-and-white photography uses this reaction to capture images in shades of gray, with the darkest areas of the film corresponding to the areas that received the most light. The first step in film processing is to enhance the black/white contrast by using a developer to increase the amount of black. The developer is a reductant: because silver atoms catalyze the reduction reaction, grains of silver bromide that have already been partially reduced by exposure to light react with the reductant much more rapidly than unexposed grains. After the film is developed, any unexposed silver bromide must be removed by a process called “fixing”; otherwise, the entire film would turn black with additional exposure to light. Although silver bromide is insoluble in water, it is soluble in a dilute solution of sodium thiosulfate (Na2S2O3; photographer’s hypo) because of the formation of [Ag(S2O3)2]3− ions. Thus Darkening of silver bromide washing the film with thiosulfate solution dissolves crystals by exposure to light. unexposed silver bromide and leaves a pattern of The top image shows AgBr before metallic silver granules that constitutes the negative. exposure to light, and the bottom This procedure is summarized in Figure 4.13 "Outline of image after exposure. the Steps Involved in Producing a Black-and-White Photograph". The negative image is then projected onto paper coated with silver halides, and the developing and fixing processes are repeated to give a positive image. (Color photography works in much the same way, with a combination of silver halides and organic dyes superimposed in layers.) “Instant photo” operations can generate more than a hundred gallons of dilute silver waste solution per day. Recovery of silver from thiosulfate fixing solutions involves first removing the thiosulfate by oxidation and then precipitating Ag+ ions with excess chloride ions.
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Figure 4.13 Outline of the Steps Involved in Producing a Black-and-White Photograph
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EXAMPLE 12 A silver recovery unit can process 1500 L of photographic silver waste solution per day. Adding excess solid sodium chloride to a 500 mL sample of the waste (after removing the thiosulfate as described previously) gives a white precipitate that, after filtration and drying, consists of 3.73 g of AgCl. What mass of NaCl must be added to the 1500 L of silver waste to ensure that all the Ag+ ions precipitate? Given: volume of solution of one reactant and mass of product from a sample of reactant solution Asked for: mass of second reactant needed for complete reaction Strategy: A Write the net ionic equation for the reaction. Calculate the number of moles of AgCl obtained from the 500 mL sample and then determine the concentration of Ag+ in the sample by dividing the number of moles of AgCl formed by the volume of solution. B Determine the total number of moles of Ag+ in the 1500 L solution by multiplying the Ag+ concentration by the total volume. C Use mole ratios to calculate the number of moles of chloride needed to react with Ag+. Obtain the mass of NaCl by multiplying the number of moles of NaCl needed by its molar mass. Solution: We can use the data provided to determine the concentration of Ag + ions in the waste, from which the number of moles of Ag+ in the entire waste solution can be calculated. From the net ionic equation, we can determine how many moles of Cl− are needed, which in turn will give us the mass of NaCl necessary. A The first step is to write the net ionic equation for the reaction: Cl−(aq) + Ag+(aq) → AgCl(s)
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We know that 500 mL of solution produced 3.73 g of AgCl. We can convert this value to the number of moles of AgCl as follows:
grams AgCl 1 mol AgCl moles AgCl = = 3.73 g AgCl molar mass AgCl 143.32 g AgCl
=
Therefore, the 500 mL sample of the solution contained 0.0260 mol of Ag +. The Ag+ concentration is determined as follows:
moles Ag + 0.0260 mol AgCl [Ag ] = = = 0.0520 M liters soln 0.500 L +
B The total number of moles of Ag+ present in 1500 L of solution is as follows:
moles Ag + = 1500 L
(
0.520 mol 1 L
)
= 78.1 mol Ag +
C According to the net ionic equation, one Cl− ion is required for each Ag+ ion. Thus 78.1 mol of NaCl are needed to precipitate the silver. The corresponding mass of NaCl is
mass NaCl = 78.1 mol NaCl
( 1 mol NaCl ) 58.44 g NaCl
= 4560 g NaCl = 4.5
Note that 78.1 mol of AgCl correspond to 8.43 kg of metallic silver, which is worth about $7983 at 2011 prices ($32.84 per troy ounce). Silver recovery may be economically attractive as well as ecologically sound, although the procedure outlined is becoming nearly obsolete for all but artistic purposes with the growth of digital photography. Exercise Because of its toxicity, arsenic is the active ingredient in many pesticides. The arsenic content of a pesticide can be measured by oxidizing arsenic compounds to the arsenate ion (AsO43−), which forms an insoluble silver salt (Ag3AsO4). Suppose you are asked to assess the purity of technical grade sodium arsenite (NaAsO2), the active ingredient in a pesticide used against
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termites. You dissolve a 10.00 g sample in water, oxidize it to arsenate, and dilute it with water to a final volume of 500 mL. You then add excess AgNO 3 solution to a 50.0 mL sample of the arsenate solution. The resulting precipitate of Ag3AsO4 has a mass of 3.24 g after drying. What is the percentage by mass of NaAsO2 in the original sample? Answer: 91.0%
Summary In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when solutions of two substances are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt.
KEY TAKEAWAY • Predicting the solubility of ionic compounds in water can give insight into whether or not a reaction will occur.
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CONCEPTUAL PROBLEMS 1. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate. a. FeCl2(aq) + Na2S(aq) b. NaOH(aq) + H3PO4(aq) c. ZnCl2(aq) + (NH4)2S(aq) 2. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate. a. KOH(aq) + H3PO4(aq) b. K2CO3(aq) + BaCl2(aq) c. Ba(NO3)2(aq) + Na2SO4(aq) 3. Which representation best corresponds to an aqueous solution originally containing each of the following? a. 1 M NH4Cl b. 1 M NaO2CCH3 c. 1 M NaOH + 1 M HCl d. 1 M Ba(OH)2 + 1 M H2SO4
4. Which representation in Problem 3 best corresponds to an aqueous solution originally containing each of the following?
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a. b. c. d.
1 M CH3CO2H + 1 M NaOH 1 M NH3 + 1 M HCl 1 M Na2CO3 + 1 M H2SO4 1 M CaCl2 + 1 M H3PO4
ANSWER 3. a. b. c. d.
1 1 1 2
NUMERICAL PROBLEMS 1. What mass of precipitate would you expect to obtain by mixing 250 mL of a solution containing 4.88 g of Na2CrO4 with 200 mL of a solution containing 3.84 g of AgNO3? What is the final nitrate ion concentration? 2. Adding 10.0 mL of a dilute solution of zinc nitrate to 246 mL of 2.00 M sodium sulfide produced 0.279 g of a precipitate. How many grams of zinc(II) nitrate and sodium sulfide were consumed to produce this quantity of product? What was the concentration of each ion in the original solutions? What is the concentration of the sulfide ion in solution after the precipitation reaction, assuming no further reaction?
ANSWER 1. 3.75 g Ag2CrO4; 5.02 × 10−2 M nitrate
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4.6 Acid–Base Reactions LEARNING OBJECTIVE 1. To know the characteristic properties of acids and bases.
Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases.
Definitions of Acids and Bases In Chapter 2 "Molecules, Ions, and Chemical Formulas", we defined acids as substances that dissolve in water to produce H+ ions, whereas bases were defined as substances that dissolve in water to produce OH− ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO3) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive.
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The Arrhenius Definition of Acids and Bases The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H+ ions (protons; Equation 4.17), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH−) ions (Equation 4.18): Equation 4.17
HCl(g)
an Arrhenius acid
H O(l)
2 ⎯⎯⎯⎯⎯ → H + (aq) + Cl− (aq)
Equation 4.18
NaOH(s)
an Arrhenius base
H O(l)
2 ⎯⎯⎯⎯⎯ → Na + (aq) + OH− (aq)
According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H+ and OH− ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substances in aqueous solution. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce H+ and OH− ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride (Equation 4.19) is not an acid–base reaction because it does not involve H+ and OH−: Equation 4.19 NH3(g) + HCl(g) → NH4Cl(s)
The Brønsted–Lowry Definition of Acids and Bases Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936),
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who defined acid–base reactions in terms of the transfer of a proton (H + ion) from one substance to another. According to Brønsted and Lowry, an acid22 is any substance that can donate a proton, and a base23 is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form NH4+, so in Equation 4.19, NH3 is a Brønsted–Lowry base and HCl is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified. We will present a third definition—Lewis acids and bases—in Chapter 8 "Ionic versus Covalent Bonding" when we discuss molecular structure.
Polyprotic Acids
22. A substance with at least one hydrogen atom that can dissociate to form an anion and + an H ion (a proton) in aqueous solution, thereby foming an acidic solution. 23. A substance that produces one or more hydroxide ions (OH− ) and a cation when dissolved in aqueous solution, thereby forming a basic solution. 24. A compound that is capable of donating one proton per molecule.
Acids differ in the number of protons they can donate. For example, monoprotic acids24 are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO3, and HNO2. All carboxylic acids that contain a single −CO2H group, such as acetic acid (CH3CO2H), are monoprotic acids, dissociating to form RCO2− and H+ (Section 4.1 "Aqueous Solutions"). Polyprotic acids25 can donate more than one proton per molecule. For example, H2SO4 can donate two H+ ions in separate steps, so it is a diprotic acid26, and H3PO4, which is capable of donating three protons in successive steps, is a triprotic acid27 (Equation 4.20, Equation 4.21, and Equation 4.22): Equation 4.20 H2 O(l)
H3 PO4 (l) ⥫=⥬ H + (aq) + H2 PO4 − (aq) Equation 4.21
H2 PO4 − (aq) ⇌ H + (aq) + HPO4 2− (aq)
25. A compound that can donate more than one proton per molecule. 26. A compound that can donate two protons per molecule in separate steps. 27. A compound that can donate three protons per molecule in separate steps.
4.6 Acid–Base Reactions
Equation 4.22
HPO4 2− (aq) ⇌ H + (aq) + PO4 3− (aq)
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28. The point at which the rates of the forward and reverse reactions become the same, so that the net composition of the system no longer changes with time. 29. An acid that reacts essentially completely with water to give H+ and the corresponding anion. 30. A base that dissociates essentially completely in water − to give OH and the corresponding cation. 31. An acid in which only a fraction of the molecules react + with water to produce H and the corresponding anion. 32. A base in which only a fraction of the molecules react with − water to produce OH and the corresponding cation.
4.6 Acid–Base Reactions
In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibrium28. We will return to the concept of equilibrium in more detail in Chapter 15 "Chemical Equilibrium".
Strengths of Acids and Bases We will not discuss the strengths of acids and bases quantitatively until Chapter 16 "Aqueous Acid–Base Equilibriums". Qualitatively, however, we can state that strong acids29 react essentially completely with water to give H+ and the corresponding anion. Similarly, strong bases30 dissociate essentially completely in water to give OH− and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids31 and weak bases32 react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form.
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In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO 3, HClO4, and H2SO4 (H3PO4 is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH)2. Common weak acids include HCN, H2S, HF, oxoacids such as HNO2 and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows: Equation 4.23 H2 O(l)
CH3 CO2 H(l) ⥫=⥬ H + (aq) + CH3 CO2 − (aq) Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates), as we stated in Section 4.1 "Aqueous Solutions". Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation 4.24) but a weak acid when it donates its second proton (Equation 4.25) as indicated by the single and double arrows, respectively: Equation 4.24 H O(l)
2 H2 SO4 (l) ⎯⎯⎯⎯⎯ → H + (aq) + HSO4 − (aq)
strong acid
Equation 4.25
HSO4 − (aq) ⇌ H + (aq) + SO4 2− (aq) weak acid
Consequently, an aqueous solution of sulfuric acid contains H +(aq) ions and a mixture of HSO4−(aq) and SO42−(aq) ions but no H2SO4 molecules. The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion: Equation 4.26
NH3 (g) + H2 O(l) ⇌ NH4 + (aq) + OH− (aq)
4.6 Acid–Base Reactions
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Most of the ammonia (>99%) is present in the form of NH 3(g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S2−). Table 4.3 "Common Strong Acids and Bases" lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or a weak electrolyte! Many weak acids and bases are extremely soluble in water.
Note the Pattern There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte.
Table 4.3 Common Strong Acids and Bases Strong Acids Hydrogen Halides
Oxoacids
Strong Bases Group 1 Hydroxides
Hydroxides of the Heavier Group 2 Elements
HCl
HNO3
LiOH
Ca(OH)2
HBr
H2SO4
NaOH
Sr(OH)2
HI
HClO4
KOH
Ba(OH)2
RbOH CsOH
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EXAMPLE 13 Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these. a. b. c. d. e.
CH3CH2CO2H CH3OH Sr(OH)2 CH3CH2NH2 HBrO4
Given: compound Asked for: acid or base strength Strategy: A Determine whether the compound is organic or inorganic. B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H+ or OH− ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H2SO4 are weak acids. Solution: a. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, so it must be a weak acid. b. A CH3OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociate to form the OH − ion. Because it does not contain a carboxylic acid (−CO 2H) group, methanol also cannot dissociate to form H+(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base. c. A Sr(OH)2 is an inorganic compound that contains one Sr2+ and two OH− ions per formula unit. B We therefore expect it to be a strong base, similar to Ca(OH)2. d. A CH3CH2NH2 is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an R group. B Consequently, we expect it to behave similarly to ammonia (Equation
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4.23), reacting with water to produce small amounts of the OH − ion. Ethylamine is therefore a weak base. e. A HBrO4 is perbromic acid, an inorganic compound. B It is not listed in Table 4.3 "Common Strong Acids and Bases" as one of the common strong acids, but that does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly below Cl in group 17. We might therefore expect that HBrO4 is chemically similar to HClO4, a strong acid—and, in fact, it is. Exercise Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these. a. b. c. d. e.
Ba(OH)2 HIO4 CH3CH2CH2CO2H (CH3)2NH CH2O
Answer: a. b. c. d. e.
strong base strong acid weak acid weak base none of these; formaldehyde is a neutral molecule
The Hydronium Ion Because isolated protons are very unstable and hence very reactive, an acid never simply “loses” an H+ ion. Instead, the proton is always transferred to another substance, which acts as a base in the Brønsted–Lowry definition. Thus in every acid–base reaction, one species acts as an acid and one species acts as a base. Occasionally, the same substance performs both roles, as you will see later. When a strong acid dissolves in water, the proton that is released is transferred to a water molecule that acts as a proton acceptor or base, as shown for the dissociation of sulfuric acid:
4.6 Acid–Base Reactions
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Equation 4.27
H2 SO4 (l) + H2 O(l) → H3 O + (aq) + HSO4 − (aq) acid
base
(proton
(proton
donor)
acceptor)
acid
base
Technically, therefore, it is imprecise to describe the dissociation of a strong acid as producing H+(aq) ions, as we have been doing. The resulting H3O+ ion, called the hydronium ion33, is a more accurate representation of H+(aq). For the sake of brevity, however, in discussing acid dissociation reactions, we will often show the product as H+(aq) (as in Equation 4.23) with the understanding that the product is actually the H3O+(aq) ion. Conversely, bases that do not contain the hydroxide ion accept a proton from water, so small amounts of OH− are produced, as in the following: Equation 4.28
NH3 (g) + H2 O(l) ⇌ NH4 + (aq) + OH− (aq) base
H+ (aq) .
+
33. The H3 O ion, represented as 34. When substances can behave as both an acid and a base. 35. The substance formed when a Brønsted–Lowry base accepts a proton. 36. The substance formed when a Brønsted–Lowry acid donates a proton. 37. An acid and a base that differ by only one hydrogen ion. All acid–base reactions involve two conjugate acid–base pairs, the Brønsted–Lowry acid and the base it forms after donating its proton, and the Brønsted–Lowry base and the acid it forms after accepting a proton.
4.6 Acid–Base Reactions
acid
acid
base
Again, the double arrow indicates that the reaction does not go to completion but rather reaches a state of equilibrium. In this reaction, water acts as an acid by donating a proton to ammonia, and ammonia acts as a base by accepting a proton from water. Thus water can act as either an acid or a base by donating a proton to a base or by accepting a proton from an acid. Substances that can behave as both an acid and a base are said to be amphoteric34. The products of an acid–base reaction are also an acid and a base. In Equation 4.27, for example, the products of the reaction are the hydronium ion, here an acid, and the hydrogen sulfate ion, here a weak base. In Equation 4.28, the products are NH4+, an acid, and OH−, a base. The product NH4+ is called the conjugate acid35 of the base NH3, and the product OH− is called the conjugate base36 of the acid H2O. Thus all acid–base reactions actually involve two conjugate acid–base pairs37; in Equation 4.28, they are NH4+/NH3 and H2O/OH−. We will describe the relationship between conjugate acid–base pairs in more detail in Chapter 16 "Aqueous Acid–Base Equilibriums".
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Neutralization Reactions A neutralization reaction38 is one in which an acid and a base react in stoichiometric amounts to produce water and a salt39, the general term for any ionic substance that does not have OH− as the anion or H+ as the cation. If the base is a metal hydroxide, then the general formula for the reaction of an acid with a base is described as follows: Acid plus base yields water plus salt. For example, the reaction of equimolar amounts of HBr and NaOH to give water and a salt (NaBr) is a neutralization reaction: Equation 4.29
HBr(aq) + NaOH(aq) → H2 O(l) + NaBr(aq) acid
base
water
salt
Note the Pattern Acid plus base yields water plus salt.
If we write the complete ionic equation for the reaction in Equation 4.29, we see that Na+(aq) and Br−(aq) are spectator ions and are not involved in the reaction: Equation 4.30
H + (aq) + Br− (aq) + Na + (aq) + OH− (aq) → H2 O(l) + Na + (aq) + The overall reaction is therefore simply the combination of H +(aq) and OH−(aq) to produce H2O, as shown in the net ionic equation: Equation 4.31 38. A chemical reaction in which an acid and a base react in stoichiometric amounts to produce water and a salt. 39. The general term for any ionic substance that does not have OH− as the anion or H+ as the cation.
4.6 Acid–Base Reactions
H+(aq) + OH–(aq) → H2O(l) The net ionic equation for the reaction of any strong acid with any strong base is identical to Equation 4.31.
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The strengths of the acid and the base generally determine whether the reaction goes to completion. The reaction of any strong acid with any strong base goes essentially to completion, as does the reaction of a strong acid with a weak base, and a weak acid with a strong base. Examples of the last two are as follows: Equation 4.32
HCl(aq) + NH3 (aq) → NH4 Cl(aq) strong acid
weak base
salt
Equation 4.33
CH3 CO2 H(aq) + NaOH(aq) → CH3 CO2 Na(aq) + H2 O(l) weak acid
strong base
salt
Sodium acetate is written with the organic component first followed by the cation, as is usual for organic salts. Most reactions of a weak acid with a weak base also go essentially to completion. One example is the reaction of acetic acid with ammonia: Equation 4.34
CH3 CO2 H(aq) + NH3 (aq) → CH3 CO2 NH4 (aq) weak acid
weak base
salt
An example of an acid–base reaction that does not go to completion is the reaction of a weak acid or a weak base with water, which is both an extremely weak acid and an extremely weak base. We will discuss these reactions in more detail in Chapter 16 "Aqueous Acid–Base Equilibriums".
Note the Pattern Except for the reaction of a weak acid or a weak base with water, acid–base reactions essentially go to completion.
In some cases, the reaction of an acid with an anion derived from a weak acid (such as HS−) produces a gas (in this case, H2S). Because the gaseous product escapes from solution in the form of bubbles, the reverse reaction cannot occur. Therefore, these reactions tend to be forced, or driven, to completion. Examples include reactions in
4.6 Acid–Base Reactions
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which an acid is added to ionic compounds that contain the HCO 3−, CN−, or S2− anions, all of which are driven to completion (Figure 4.14 "The Reaction of Dilute Aqueous HNO"): Equation 4.35
HCO3 − (aq) + H + (aq) → H2 CO3 (aq) H2 CO3 (aq) → CO2 (g) + H2 O(l)
Equation 4.36
CN− (aq) + H + (aq) → HCN(g) Equation 4.37
S2− (aq) + H + (aq) → HS− (aq) HS− (aq) + H + (aq) → H2 S(g)
The reactions in Equation 4.37 are responsible for the rotten egg smell that is produced when metal sulfides come in contact with acids.
Figure 4.14 The Reaction of Dilute Aqueous HNO3 with a Solution of Na2CO3
Note the vigorous formation of gaseous CO2.
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EXAMPLE 14 Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH3CH2CO2H) with aqueous calcium hydroxide [Ca(OH)2] to give calcium propionate. Do you expect this reaction to go to completion, making it a feasible method for the preparation of calcium propionate? Given: reactants and product Asked for: balanced chemical equation and whether the reaction will go to completion Strategy: Write the balanced chemical equation for the reaction of propionic acid with calcium hydroxide. Based on their acid and base strengths, predict whether the reaction will go to completion. Solution: Propionic acid is an organic compound that is a weak acid, and calcium hydroxide is an inorganic compound that is a strong base. The balanced chemical equation is as follows: 2CH3CH2CO2H(aq) + Ca(OH)2(aq) → (CH3CH2CO2)2Ca(aq) + 2H2O(l) The reaction of a weak acid and a strong base will go to completion, so it is reasonable to prepare calcium propionate by mixing solutions of propionic acid and calcium hydroxide in a 2:1 mole ratio. Exercise Write a balanced chemical equation for the reaction of solid sodium acetate with dilute sulfuric acid to give sodium sulfate. Answer: 2CH3CO2Na(s) + H2SO4(aq) → Na2SO4(aq) + 2CH3CO2H(aq)
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One of the most familiar and most heavily advertised applications of acid–base chemistry is antacids, which are bases that neutralize stomach acid. The human stomach contains an approximately 0.1 M solution of hydrochloric acid that helps digest foods. If the protective lining of the stomach breaks down, this acid can attack the stomach tissue, resulting in the formation of an ulcer. Because one factor that is believed to contribute to the formation of stomach ulcers is the production of excess acid in the stomach, many individuals routinely consume large quantities of antacids. The active ingredients in antacids include Stomach acid. An antacid tablet sodium bicarbonate and potassium bicarbonate reacts with 0.1 M HCl (the (NaHCO3 and KHCO3; Alka-Seltzer); a mixture of approximate concentration found in the human stomach). magnesium hydroxide and aluminum hydroxide [Mg(OH)2 and Al(OH)3; Maalox, Mylanta]; calcium carbonate (CaCO3; Tums); and a complex salt, dihydroxyaluminum sodium carbonate [NaAl(OH)2CO3; original Rolaids]. Each has certain advantages and disadvantages. For example, Mg(OH)2 is a powerful laxative (it is the active ingredient in milk of magnesia), whereas Al(OH)3 causes constipation. When mixed, each tends to counteract the unwanted effects of the other. Although all antacids contain both an anionic base (OH−, CO32−, or HCO3−) and an appropriate cation, they differ substantially in the amount of active ingredient in a given mass of product.
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EXAMPLE 15 Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO 3? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.) Given: volume and molarity of acid and mass of base in an antacid tablet Asked for: number of tablets required for 90% neutralization Strategy: A Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion. B Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid. C Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass. Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet. Solution: A We first write the balanced chemical equation for the reaction: 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2CO3(aq) Each carbonate ion can react with 2 mol of H+ to produce H2CO3, which rapidly decomposes to H2O and CO2. Because HCl is a strong acid and CO32− is a weak base, the reaction will go to completion. B Next we need to determine the number of moles of HCl present:
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75 mL
1 L
( 1000 mL ) (
0.20 mol HCl L
)
= 0.015 mol HCl
Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90: (0.015 mol HCl)(0.90) = 0.014 mol HCl We know from the stoichiometry of the reaction that each mole of CaCO 3 reacts with 2 mol of HCl, so we need
moles CaCO3 = 0.014 mol HCl C Each Tums tablet contains
500 mg CaCO 3 1 Tums tablet Thus we need
( 2 mol HCl ) 1 mol CaCO 3
1 g 1000 mg CaCO3
0.0070 mol CaCO 3 0.00500 mol CaCO 3
= 0.0070 mol CaC
1 mol CaCO 3 = 0.00500 ( 100.1 g )
= 1.4 Tums tablets.
Exercise Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH) 2CO3? The neutralization reaction can be written as follows: NaAl(OH)2CO3(s) + 4HCl(aq) → AlCl3(aq) + NaCl(aq) + CO2(g) + 3H2O(l) Answer: 6.4 tablets
40. A logarithmic scale used to express the hydrogen ion (H+ ) concentration of a solution, making it possible to describe acidity or basicity quantitatively.
4.6 Acid–Base Reactions
The pH Scale One of the key factors affecting reactions that occur in dilute solutions of acids and bases is the concentration of H+ and OH− ions. The pH scale40 provides a convenient
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way of expressing the hydrogen ion (H+) concentration of a solution and enables us to describe acidity or basicity in quantitative terms. Pure liquid water contains extremely low but measurable concentrations of H3O+(aq) and OH−(aq) ions produced via an autoionization reaction, in which water acts simultaneously as an acid and as a base: Equation 4.38 H2O(l) + H2O(l) ? H3O+(aq) + OH−(aq) The concentration of hydrogen ions in pure water is only 1.0 × 10−7 M at 25°C. Because the autoionization reaction produces both a proton and a hydroxide ion, the OH− concentration in pure water is also 1.0 × 10−7 M. Pure water is a neutral solution41, in which [H+] = [OH−] = 1.0 × 10−7 M. The pH scale describes the hydrogen ion concentration of a solution in a way that avoids the use of exponential notation; pH42 is defined as the negative base-10 logarithm of the hydrogen ion concentration:pH is actually defined as the negative base-10 logarithm of hydrogen ion activity. As you will learn in a more advanced course, the activity of a substance in solution is related to its concentration. For dilute solutions such as those we are discussing, the activity and the concentration are approximately the same. Equation 4.39 pH = −log[H+] Conversely, Equation 4.40 [H+] = 10−pH 41. A solution in which the total positive charge from all the cations is matched by an identical total negative charge from all the anions. 42. The negative base-10 logarithm of the hydrogen ion concentration:
(If you are not familiar with logarithms or using a calculator to obtain logarithms and antilogarithms, consult Essential Skills 3 in Section 4.1 "Aqueous Solutions"0.) Because the hydrogen ion concentration is 1.0 × 10−7 M in pure water at 25°C, the pH of pure liquid water (and, by extension, of any neutral solution) is
pH = −log[H+ ].
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Equation 4.41 pH = −log[1.0 × 10−7] = 7.00 Adding an acid to pure water increases the hydrogen ion concentration and decreases the hydroxide ion concentration because a neutralization reaction occurs, such as that shown in Equation 4.31. Because the negative exponent of [H+] becomes smaller as [H+] increases, the pH decreases with increasing [H+]. For example, a 1.0 M solution of a strong monoprotic acid such as HCl or HNO 3 has a pH of 0.00: Equation 4.42 pH = −log[1.0] = 0.00
Note the Pattern pH decreases with increasing [H+].
Conversely, adding a base to pure water increases the hydroxide ion concentration and decreases the hydrogen ion concentration. Because the autoionization reaction of water does not go to completion, neither does the neutralization reaction. Even a strongly basic solution contains a detectable amount of H + ions. For example, a 1.0 M OH− solution has [H+] = 1.0 × 10−14 M. The pH of a 1.0 M NaOH solution is therefore Equation 4.43 pH = −log[1.0 × 10−14] = 14.00 For practical purposes, the pH scale runs from pH = 0 (corresponding to 1 M H +) to pH 14 (corresponding to 1 M OH−), although pH values less than 0 or greater than 14 are possible. We can summarize the relationships between acidity, basicity, and pH as follows: • If pH = 7.0, the solution is neutral. • If pH < 7.0, the solution is acidic.
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• If pH > 7.0, the solution is basic. Keep in mind that the pH scale is logarithmic, so a change of 1.0 in the pH of a solution corresponds to a tenfold change in the hydrogen ion concentration. The foods and consumer products we encounter daily represent a wide range of pH values, as shown in Figure 4.15 "A Plot of pH versus [H". Figure 4.15 A Plot of pH versus [H+] for Some Common Aqueous Solutions
Although many substances exist in a range of pH values (indicated in parentheses), they are plotted using typical values.
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EXAMPLE 16 a. What is the pH of a 2.1 × 10−2 M aqueous solution of HClO4? b. The pH of a vinegar sample is 3.80. What is its hydrogen ion concentration? Given: molarity of acid or pH Asked for: pH or [H+] Strategy: Using the balanced chemical equation for the acid dissociation reaction and Equation 4.41 or 4.40, determine [H+] and convert it to pH or vice versa. Solution:
a. HClO4 (perchloric acid) is a strong acid, so it dissociates completely into H+ ions and ClO4− ions: HClO4(l) → H+(aq) + ClO4−(aq) The H+ ion concentration is therefore the same as the perchloric acid concentration. The pH of the perchloric acid solution is thus pH = −log[H+] = −log(2.1 × 10−2) = 1.68 The result makes sense: the H+ ion concentration is between 10−1 M and 10−2 M, so the pH must be between 1 and 2. Note: The assumption that [H+] is the same as the concentration of the acid is valid for only strong acids. Because weak acids do not dissociate completely in aqueous solution, a more complex procedure is needed to calculate the pH of their solutions, which we will describe in Chapter 16 "Aqueous Acid–Base Equilibriums". b. We are given the pH and asked to calculate the hydrogen ion concentration. From Equation 4.40,
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10−pH = [H+] Thus [H+] = 10−3.80 = 1.6 × 10−4 M. Exercise a. What is the pH of a 3.0 × 10−5 M aqueous solution of HNO3? b. What is the hydrogen ion concentration of turnip juice, which has a pH of 5.41? Answer: a. pH = 4.52 b. [H+] = 3.9 × 10−6 M
Tools have been developed that make the measurement of pH simple and convenient (Figure 4.16 "Two Ways of Measuring the pH of a Solution: pH Paper and a pH Meter"). For example, pH paper consists of strips of paper impregnated with one or more acid–base indicators43, which are intensely colored organic molecules whose colors change dramatically depending on the pH of the solution. Placing a drop of a solution on a strip of pH paper and comparing its color with standards give the solution’s approximate pH. A more accurate tool, the pH meter, uses a glass electrode, a device whose voltage depends on the H+ ion concentration.
43. An intensely colored organic molecule whose color changes dramatically depending on the pH of the solution.
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Figure 4.16 Two Ways of Measuring the pH of a Solution: pH Paper and a pH Meter
Note that both show that the pH is 1.7, but the pH meter gives a more precise value.
KEY EQUATIONS definition of pH Equation 4.39: pH = −log[H+] Equation 4.40: [H+] = 10−pH
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Summary Acid–base reactions require both an acid and a base. In Brønsted–Lowry terms, an acid is a substance that can donate a proton (H+), and a base is a substance that can accept a proton. All acid–base reactions contain two acid–base pairs: the reactants and the products. Acids can donate one proton (monoprotic acids), two protons (diprotic acids), or three protons (triprotic acids). Compounds that are capable of donating more than one proton are generally called polyprotic acids. Acids also differ in their tendency to donate a proton, a measure of their acid strength. Strong acids react completely with water to produce H3O+(aq) (the hydronium ion), whereas weak acids dissociate only partially in water. Conversely, strong bases react completely with water to produce the hydroxide ion, whereas weak bases react only partially with water to form hydroxide ions. The reaction of a strong acid with a strong base is a neutralization reaction, which produces water plus a salt. The acidity or basicity of an aqueous solution is described quantitatively using the pH scale. The pH of a solution is the negative logarithm of the H+ ion concentration and typically ranges from 0 for strongly acidic solutions to 14 for strongly basic ones. Because of the autoionization reaction of water, which produces small amounts of hydronium ions and hydroxide ions, a neutral solution of water contains 1 × 10−7 M H+ ions and has a pH of 7.0. An indicator is an intensely colored organic substance whose color is pH dependent; it is used to determine the pH of a solution.
KEY TAKEAWAY • An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt.
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CONCEPTUAL PROBLEMS 1. Why was it necessary to expand on the Arrhenius definition of an acid and a base? What specific point does the Brønsted–Lowry definition address? 2. State whether each compound is an acid, a base, or a salt. a. b. c. d. e.
CaCO3 NaHCO3 H2SO4 CaCl2 Ba(OH)2
3. State whether each compound is an acid, a base, or a salt. a. b. c. d. e.
NH3 NH4Cl H2CO3 CH3COOH NaOH
4. Classify each compound as a strong acid, a weak acid, a strong base, or a weak base in aqueous solution. a. b. c. d. e. f. g. h.
sodium hydroxide acetic acid magnesium hydroxide tartaric acid sulfuric acid ammonia hydroxylamine (NH2OH) hydrocyanic acid
5. Decide whether each compound forms an aqueous solution that is strongly acidic, weakly acidic, strongly basic, or weakly basic. a. b. c. d. e. f. g. h.
4.6 Acid–Base Reactions
propanoic acid hydrobromic acid methylamine lithium hydroxide citric acid sodium acetate ammonium chloride barium hydroxide
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6. What is the relationship between the strength of an acid and the strength of the conjugate base derived from that acid? Would you expect the CH 3CO2− ion to be a strong base or a weak base? Why? Is the hydronium ion a strong acid or a weak acid? Explain your answer. 7. What are the products of an acid–base reaction? Under what circumstances is one of the products a gas? 8. Explain how an aqueous solution that is strongly basic can have a pH, which is a measure of the acidity of a solution.
ANSWER 5. a. b. c. d. e. f. g. h.
4.6 Acid–Base Reactions
weakly acidic strongly acidic weakly basic strongly basic weakly acidic weakly basic weakly acidic strongly basic
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 3 (Section 4.1 "Aqueous Solutions"0) before proceeding to the Numerical Problems. 1. Derive an equation to relate the hydrogen ion concentration to the molarity of a solution of a strong monoprotic acid. 2. Derive an equation to relate the hydroxide ion concentration to the molarity of a solution of a. a group I hydroxide. b. a group II hydroxide. 3. Given the following salts, identify the acid and the base in the neutralization reactions and then write the complete ionic equation: a. b. c. d.
barium sulfate lithium nitrate sodium bromide calcium perchlorate
4. What is the pH of each solution? a. b. c. d.
5.8 × 10−3 mol of HNO3 in 257 mL of water 0.0079 mol of HI in 750 mL of water 0.011 mol of HClO4 in 500 mL of water 0.257 mol of HBr in 5.00 L of water
5. What is the hydrogen ion concentration of each substance in the indicated pH range? a. black coffee (pH 5.10) b. milk (pH 6.30–7.60) c. tomatoes (pH 4.00–4.40) 6. What is the hydrogen ion concentration of each substance in the indicated pH range? a. orange juice (pH 3–4) b. fresh egg white (pH 7.60–7.80) c. lemon juice (pH 2.20–2.40) 7. What is the pH of a solution prepared by diluting 25.00 mL of 0.879 M HCl to a volume of 555 mL?
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8. Vinegar is primarily an aqueous solution of acetic acid. Commercial vinegar typically contains 5.0 g of acetic acid in 95.0 g of water. What is the concentration of commercial vinegar? If only 3.1% of the acetic acid dissociates to CH3CO2− and H+, what is the pH of the solution? (Assume the density of the solution is 1.00 g/mL.) 9. If a typical household cleanser is 0.50 M in strong base, what volume of 0.998 M strong monoprotic acid is needed to neutralize 50.0 mL of the cleanser? 10. A 25.00 mL sample of a 0.9005 M solution of HCl is diluted to 500.0 mL. What is the molarity of the final solution? How many milliliters of 0.223 M NaOH are needed to neutralize 25.00 mL of this final solution? 11. If 20.0 mL of 0.10 M NaOH are needed to neutralize 15.0 mL of gastric fluid, what is the molarity of HCl in the fluid? (Assume all the acidity is due to the presence of HCl.) What other base might be used instead of NaOH? 12. Malonic acid (C3H4O4) is a diprotic acid used in the manufacture of barbiturates. How many grams of malonic acid are in a 25.00 mL sample that requires 32.68 mL of 1.124 M KOH for complete neutralization to occur? Malonic acid is a dicarboxylic acid; propose a structure for malonic acid. 13. Describe how you would prepare 500 mL of a 1.00 M stock solution of HCl from an HCl solution that is 12.11 M. Using your stock solution, how would you prepare 500 mL of a solution that is 0.012 M in HCl? 14. Given a stock solution that is 8.52 M in HBr, describe how you would prepare a 500 mL solution with each concentration. a. 2.50 M b. 4.00 × 10−3 M c. 0.989 M 15. How many moles of solute are contained in each? a. 25.00 mL of 1.86 M NaOH b. 50.00 mL of 0.0898 M HCl c. 13.89 mL of 0.102 M HBr 16. A chemist needed a solution that was approximately 0.5 M in HCl but could measure only 10.00 mL samples into a 50.00 mL volumetric flask. Propose a method for preparing the solution. (Assume that concentrated HCl is 12.0 M.) 17. Write the balanced chemical equation for each reaction. a. perchloric acid with potassium hydroxide b. nitric acid with calcium hydroxide
4.6 Acid–Base Reactions
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18. Write the balanced chemical equation for each reaction. a. solid strontium hydroxide with hydrobromic acid b. aqueous sulfuric acid with solid sodium hydroxide 19. A neutralization reaction gives calcium nitrate as one of the two products. Identify the acid and the base in this reaction. What is the second product? If the product had been cesium iodide, what would have been the acid and the base? What is the complete ionic equation for each reaction?
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ANSWERS 1. [H3O+] = [HA] M 3. a. b. c. d.
H2SO4 and Ba(OH)2; 2H+ + SO42− + Ba2+ + 2OH− → 2H2O + Ba2+ + SO42− HNO3 and LiOH; H+ + NO3− + Li+ + OH− → H2O + Li+ + NO3− HBr and NaOH; H+ + Br− + Na+ + OH− → H2O + Na+ + Br− HClO4 and Ca(OH)2; 2H+ + 2ClO4− + Ca2+ + 2OH− → 2H2O + Ca2+ + 2ClO4−
5. a. 7.9 × 10−6 M H+ b. 5.0 × 10−7 to 2.5 × 10−8 M H+ c. 1.0 × 10−4 to 4.0 × 10−5 M H+ 7. pH = 1.402 9. 25 mL 11. 0.13 M HCl; magnesium carbonate, MgCO3, or aluminum hydroxide, Al(OH)3 13. 1.00 M solution: dilute 41.20 mL of the concentrated solution to a final volume of 500 mL. 0.012 M solution: dilute 12.0 mL of the 1.00 M stock solution to a final volume of 500 mL. 15. a. 4.65 × 10−2 mol NaOH b. 4.49 × 10−3 mol HCl c. 1.42 × 10−3 mol HBr 17. a. HClO4 + KOH → KClO4 + H2O b. 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O 19. The acid is nitric acid, and the base is calcium hydroxide. The other product is water. 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O The acid is hydroiodic acid, and the base is cesium hydroxide. The other product is water. HI + CsOH → CsI + H2O The complete ionic equations are 2H+ + 2NO3− + Ca2+ + 2OH− → Ca2+ + 2NO3− + H2O H+ + I− + Cs+ + OH− → Cs+ + I− + H2O
4.6 Acid–Base Reactions
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4.7 The Chemistry of Acid Rain LEARNING OBJECTIVE 1. To understand the chemistry of acid rain.
Acid–base reactions can have a strong environmental impact. For example, a dramatic increase in the acidity of rain and snow over the past 150 years is dissolving marble and limestone surfaces, accelerating the corrosion of metal objects, and decreasing the pH of natural waters. This environmental problem is called acid rain44 and has significant consequences for all living organisms. To understand acid rain requires an understanding of acid–base reactions in aqueous solution. The term acid rain is actually somewhat misleading because even pure rainwater collected in areas remote from civilization is slightly acidic (pH ≈ 5.6) due to dissolved carbon dioxide, which reacts with water to give carbonic acid, a weak acid: Equation 4.44 CO2(g) + H2O(l) ? H2CO3(aq) ? H+(aq) + HCO3−(aq) The English chemist Robert Angus Smith is generally credited with coining the phrase acid rain in 1872 to describe the increased acidity of the rain in British industrial centers (such as Manchester), which was apparently caused by the unbridled excesses of the early Industrial Revolution, although the connection was not yet understood. At that time, there was no good way to measure hydrogen ion concentrations, so it is difficult to know the actual pH of the rain observed by Smith. Typical pH values for rain in the continental United States now range from 4 to 4.5, with values as low as 2.0 reported for areas such as Los Angeles. Recall from Figure 4.15 "A Plot of pH versus [H" that rain with a pH of 2 is comparable in acidity to lemon juice, and even “normal” rain is now as acidic as tomato juice or black coffee.
44. Precipitation that is dramatically more acidic because of human activities.
What is the source of the increased acidity in rain and snow? Chemical analysis shows the presence of large quantities of sulfate (SO42−) and nitrate (NO3−) ions, and a wide variety of evidence indicates that a significant fraction of these species come
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from nitrogen and sulfur oxides produced during the combustion of fossil fuels. At the high temperatures found in both internal combustion engines and lightning discharges, molecular nitrogen and molecular oxygen react to give nitric oxide: Equation 4.45 N2(g) + O2(g) → 2NO(g) Nitric oxide then reacts rapidly with excess oxygen to give nitrogen dioxide, the compound responsible for the brown color of smog: Equation 4.46 2NO(g) + O2(g) → 2NO2(g) When nitrogen dioxide dissolves in water, it forms a 1:1 mixture of nitrous acid and nitric acid: Equation 4.47 2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq) Because molecular oxygen eventually oxidizes nitrous acid to nitric acid, the overall reaction is Equation 4.48 2N2(g) + 5O2(g) + 2H2O(l) → 4HNO3(aq) Large amounts of sulfur dioxide have always been released into the atmosphere by natural sources, such as volcanoes, forest fires, and the microbial decay of organic materials, but for most of Earth’s recorded history the natural cycling of sulfur from the atmosphere into oceans and rocks kept the acidity of rain and snow in check. Unfortunately, the burning of fossil fuels seems to have tipped the balance. Many coals contain as much as 5%–6% pyrite (FeS2) by mass, and fuel oils typically contain at least 0.5% sulfur by mass. Since the mid-19th century, these fuels have been burned on a huge scale to supply the energy needs of our modern industrial society, releasing tens of millions of tons of additional SO 2 into the atmosphere annually. In addition, roasting sulfide ores to obtain metals such as zinc and copper produces large amounts of SO2 via reactions such as
4.7 The Chemistry of Acid Rain
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Equation 4.49 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) Regardless of the source, the SO2 dissolves in rainwater to give sulfurous acid (Equation 4.50), which is eventually oxidized by oxygen to sulfuric acid (Equation 4.51): Equation 4.50 SO2(g) + H2O(l) → H2SO3(aq) Equation 4.51 2H2SO3(aq) + O2(g) → 2H2SO4(aq) Concerns about the harmful effects of acid rain have led to strong pressure on industry to minimize the release of SO2 and NO. For example, coal-burning power plants now use SO2 “scrubbers,” which trap SO2 by its reaction with lime (CaO) to produce calcium sulfite dihydrate (CaSO3·2H2O; Figure 4.17 "Schematic Diagram of a Wet Scrubber System"). Figure 4.17 Schematic Diagram of a Wet Scrubber System
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In coal-burning power plants, SO2 can be removed (“scrubbed”) from exhaust gases by its reaction with a lime (CaO) and water spray to produce calcium sulfite dihydrate (CaSO 3·2H2O). Removing SO2 from the gases prevents its conversion to SO3 and subsequent reaction with rainwater (acid rain). Scrubbing systems are now commonly used to minimize the environmental effects of large-scale fossil fuel combustion.
The damage that acid rain does to limestone and marble buildings and sculptures is due to a classic acid–base reaction. Marble and limestone both consist of calcium carbonate (CaCO3), a salt derived from the weak acid H2CO3. As we saw in Section 4.6 "Acid–Base Reactions", the reaction of a strong acid with a salt of a weak acid goes to completion. Thus we can write the reaction of limestone or marble with dilute sulfuric acid as follows: Equation 4.52 CaCO3(s) + H2SO4(aq) → CaSO4(s) + H2O(l) + CO2(g) Because CaSO4 is sparingly soluble in water, the net result of this reaction is to dissolve the marble or limestone. The Lincoln Memorial in Washington, DC, which was built in 1922, already shows significant damage from acid rain, and many older objects are exhibiting even greater damage (Figure 4.18 "Acid Rain Damage to a Statue of George Washington"). Metal objects can also suffer damage from acid rain through oxidation–reduction reactions, which are discussed in Section 4.8 "Oxidation–Reduction Reactions in Solution". The biological effects of acid rain are more complex. As indicated in Figure 4.15 "A Plot of pH versus [H", biological fluids, such as blood, have a pH of 7–8. Organisms such as fish can maintain their internal pH in water that has a pH in the range of 6.5–8.5. If the external pH is too low, however, many aquatic organisms can no longer maintain their internal pH, so they die. A pH of 4 or lower is fatal for virtually all fish, most invertebrate animals, and many microorganisms. As a result of acid rain, the pH of some lakes in Europe and the United States has dropped below 4. Recent surveys suggest that up to 6% of the lakes in the Adirondack Mountains of upstate New York and 4% of the lakes in Sweden and Norway are essentially dead and contain no fish. Neither location contains large concentrations of industry, but New York lies downwind of the industrial Midwest, and Scandinavia is downwind
4.7 The Chemistry of Acid Rain
Figure 4.18 Acid Rain Damage to a Statue of George Washington
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of the most industrialized regions of western Europe. Both regions appear to have borne the brunt of the pollution produced by their upwind neighbors. One possible way to counter the effects of acid rain in isolated lakes is by adding large quantities of finely ground limestone, which neutralizes the acid via the reaction shown in Equation 4.52 (see Section 4.1 "Aqueous Solutions"1, Problem 15).
Both marble and limestone consist of CaCO3, which reacts with acid rain in an acid–base reaction to produce CaSO4. Because CaSO4 is somewhat soluble in water, significant damage to the structure can result.
A second major way in which acid rain can cause biological damage is less direct. Trees and many other plants are sensitive to the presence of aluminum and other metals in groundwater. Under normal circumstances, aluminum hydroxide [Al(OH) 3], which is present in some soils, is insoluble. At lower pH values, however, Al(OH) 3 dissolves via the following reaction: Equation 4.53 Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3H2O(l) The result is increased levels of Al3+ ions in groundwater. Because the Al3+ ion is toxic to plants, high concentrations can affect plant growth. Acid rain can also weaken the leaves and roots of plants so much that the plants are unable to withstand other stresses. The combination of the two effects can cause significant damage to established forests, such as the Black Forest in Germany and the forests of the northeastern United States and Canada and other countries (Figure 4.19 "Acid Rain Damage to a Forest in the Czech Republic").
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Figure 4.19 Acid Rain Damage to a Forest in the Czech Republic
Trees and many other plants are sensitive to aluminum and other metals in groundwater. Acid rain increases the concentration of Al3+ in groundwater, thereby adversely affecting plant growth. Large sections of established forests have been severely damaged.
Summary Acid rain is rainfall whose pH is less than 5.6, the value typically observed, due to the presence of dissolved carbon dioxide. Acid rain is caused by nitrogen oxides and sulfur dioxide produced by both natural processes and the combustion of fossil fuels. Eventually, these oxides react with oxygen and water to give nitric acid and sulfuric acid.
KEY TAKEAWAY • The damaging effects of acid rain have led to strong pressure on industry to minimize the release of harmful reactants.
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CONCEPTUAL PROBLEMS 1. Why is it recommended that marble countertops not be used in kitchens? Marble is composed mostly of CaCO3. 2. Explain why desulfurization of fossil fuels is an area of intense research. 3. What is the role of NOx in the formation of acid rain?
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4.8 Oxidation–Reduction Reactions in Solution LEARNING OBJECTIVE 1. To identify oxidation–reduction reactions in solution.
We described the defining characteristics of oxidation–reduction, or redox, reactions in Chapter 3 "Chemical Reactions". Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reaction method. We show you how to balance redox equations using oxidation states in this section; the halfreaction method is described in Chapter 19 "Electrochemistry".
Balancing Redox Equations Using Oxidation States To balance a redox equation using the oxidation state method45, we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr2+(aq) with manganese dioxide (MnO2) in the presence of dilute acid. Equation 4.54 is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described in Section 3.5 "Classifying Chemical Reactions": Equation 4.54 +2 2+
+1 +
+4
+3 3+
+2 2+
+1
Cr (aq) + Mn O2 (s) + H (aq) → Cr (aq) + Mn (aq) + H2 O (l) –2
45. A procedure for balancing oxidation–reduction (redox) reactions in which the overall reaction is conceptually separated into two parts: an oxidation and a reduction.
−2
Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced:
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Equation 4.55 Cr2+ + Mn4+ → Cr3+ + Mn2+ The oxidation can be written as Equation 4.56 Cr2+ → Cr3+ + e− and the reduction as Equation 4.57 Mn4+ + 2e− → Mn2+ For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation equation by 2 to give Equation 4.58 2Cr2+ → 2Cr3+ + 2e−
Note the Pattern In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction:
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Equation 4.59
2Cr 2+ → 2Cr 3+ + 2e– Mn 4+ + 2e– → Mn 2+
We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products: Equation 4.60
2Cr 2+ (aq)
⎯→
2Cr 3+ (aq) + 2e−
MnO 2 (s) + 2e−
⎯→
Mn 2 + (aq)
2Cr 2 + (aq) + MnO 2 (s) ⎯→ 2Cr 3 + (aq) + Mn 2 + (aq) Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of the equation (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add H + as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding OH− as necessary to either side of the equation to balance the charges. In this case, adding four H+ ions to the left side of the equation gives Equation 4.61 2Cr2+(aq) + MnO2(s) + 4H+(aq) → 2Cr3+(aq) + Mn2+(aq) Although the charges are now balanced, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding H2O as necessary to either side of the equation. Here, we need to add two H2O molecules to the right side: Equation 4.62 2Cr2+(aq) + MnO2(s) + 4H+(aq) → 2Cr3+(aq) + Mn2+(aq) + 2H2O(l) Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing
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reactions is summarized in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" and illustrated in Example 17. Table 4.4 Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method
1. Write the unbalanced chemical equation for the reaction, showing the reactants and the products. 2. Assign oxidation states to all atoms in the reactants and the products (see Section 3.5 "Classifying Chemical Reactions") and determine which atoms change oxidation state. 3. Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number of electrons accepted or donated by each. 4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. 5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms in step 4. 6. Add the two equations and cancel the electrons. 7. Balance the charge by adding H+ or OH− ions as necessary for reactions in acidic or basic solution, respectively. 8. Balance the oxygen atoms by adding H2O molecules to one side of the equation. 9. Check to make sure that the equation is balanced in both atoms and total charges.
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EXAMPLE 17 Arsenic acid (H3AsO4) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH3, a highly toxic and unstable gas) and Zn2+(aq). Balance the equation for this reaction using oxidation states: H3AsO4(aq) + Zn(s) → AsH3(g) + Zn2+(aq) Given: reactants and products in acidic solution Asked for: balanced chemical equation using oxidation states Strategy: Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced. Solution: 1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step. 2. Assign oxidation states using the procedure described in Section 3.5 "Classifying Chemical Reactions" and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in Zn2+(aq) is +2: +5
+2 2+
0
H 3 AsO 4 (aq) + Zn (s) → AsH 3 (g) + Zn –3
(aq)
3. Write separate equations for oxidation and reduction. The arsenic atom in H3AsO4 is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:
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Reduction: As 5+ + 8e− → As3– –3
Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:
Oxidation: Zn → Zn 2+ + 2e− 4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain Reduction (× 1): As5+ + 8e– → As3– Oxidation (× 4): 4Zn0 → 4Zn2+ + 8e– 5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives Reduction: H3AsO4(aq) + 8e– → AsH3(g) Oxidation: 4Zn(s) → 4Zn2+(aq) + 8e– 6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is
H 3 AsO 4 (aq) + 4Zn(s) + 8e– → AsH 3 (g) + 4Zn 2+ (aq) + 8e– which then yields H3AsO4(aq) + 4Zn(s) → AsH3(g) + 4Zn2+(aq) 7. Balance the charge by adding H+or OH−ions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H+ ions to whichever side of the equation requires them to balance the charge. The overall charge
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on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H+ ions to the left side gives a charge of +8 on both sides of the equation: H3AsO4(aq) + 4Zn(s) + 8H+(aq) → AsH3(g) + 4Zn2+(aq) 8. Balance the oxygen atoms by adding H2O molecules to one side of the equation. There are 4 O atoms on the left side of the equation. Adding 4 H2O molecules to the right side balances the O atoms: H3AsO4(aq) + 4Zn(s) + 8H+(aq) → AsH3(g) + 4Zn2+(aq) + 4H2O(l) Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms. 9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation: Atoms: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H Total charge: 8(+1) = 4(+2) = +8 The balanced chemical equation for the reaction is therefore: H3AsO4(aq) + 4Zn(s) + 8H+(aq) → AsH3(g) + 4Zn2+(aq) + 4H2O(l) Exercise Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states: CuS(s) + H+(aq) + NO3−(aq) → Cu2+(aq) + NO(g) + SO42−(aq) Answer: 3CuS(s) + 8H+(aq) + 8NO3−(aq) → 3Cu2+(aq) + 8NO(g) + 3SO42−(aq) + 4H2O(l)
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Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example 18.
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EXAMPLE 18 The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows: Al(s) + H2O(aq) → [Al(OH)4]−(aq) + H2(g) Balance this equation using oxidation states. Given: reactants and products in a basic solution Asked for: balanced chemical equation Strategy: Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced. Solution: We will apply the same procedure used in Example 17 but in a more abbreviated form. 1. The equation for the reaction is given, so we can skip this step. 2. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH)4]−. The oxidation state of hydrogen changes from +1 in H2O to 0 in H2. Aluminum is oxidized, while hydrogen is reduced: 0
+1
+3
–
0
Al(s) + H 2 O(aq) → [Al(OH) 4 ] (aq) + H 2 (g) 3.
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Reduction: Al 0 → Al3+ + 3e− Oxidation: H
+
+ e− → H 0 (in H 2 )
4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation: Reduction: 3H+ + 3e– → 3H0 (in H2) Oxidation: Al0 → Al3+ + 3e– 5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of H 2O contains two protons, in this case, 3H+ corresponds to 3/2H2O. Similarly, each molecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H2.
3 3 H 2 O + 3e− → H 2 2 2 – Oxidation: Al → [Al(OH) 4 ] + 3e– Reduction:
6. Adding the equations and canceling the electrons gives
Al +
3 3 H 2 O + 3e− → [Al(OH) 4 ]− + H 2 + 3e− 2 2 3 3 Al + H 2 O → [Al(OH) 4 ]− + H 2 2 2
To remove the fractional coefficients, multiply both sides of the equation by 2: 2Al + 3H2O → 2[Al(OH)4]– + 3H2 7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two OH− ions to the left side:
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2Al + 2OH– + 3H2O → 2[Al(OH)4]– + 3H2 8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H2O molecules to the left side: 2Al + 2OH– + 6H2O → 2[Al(OH)4]– + 3H2 9. Be sure the equation is balanced: Atoms: 2Al + 8O + 14H = 2Al + 8O + 14H Total charge: (2)(0) + (2)(–1) + (6)(0) = (2)(–1) + (3)(0) −2 = −2 The balanced chemical equation is therefore 2Al(s) + 2OH–(aq) + 6H2O(l) → 2[Al(OH)4]–(aq) + 3H2(g) Thus 3 mol of H2 gas are produced for every 2 mol of Al. Exercise The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction. Answer: 2MnO4−(aq) + 3NO2−(aq) + H2O(l) → 2MnO2(s) + 3NO3−(aq) + 2OH−(aq)
As suggested in Example 17 and Example 18, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:
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1. Compounds of elements in high oxidation states (such as ClO 4−, NO3−, MnO4−, Cr2O72−, and UF6) tend to act as oxidants and become reduced in chemical reactions. 2. Compounds of elements in low oxidation states (such as CH 4, NH3, H2S, and HI) tend to act as reductants and become oxidized in chemical reactions.
Note the Pattern Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.
When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.
Redox Reactions of Solid Metals in Aqueous Solution A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure 4.20 "Rust Formation"). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl− ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas: Equation 4.63 Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) In subsequent steps, FeCl2 undergoes oxidation to form a reddish-brown precipitate of Fe(OH)3.
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Many metals dissolve through reactions of this type, which have the general form
Figure 4.20 Rust Formation
Equation 4.64 metal + acid → salt + hydrogen Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves: Equation 4.65 Pb(s) + 2H+(aq) → Pb2+(aq) + H2(g)
The corrosion process involves an oxidation–reduction reaction in which metallic iron is converted to Fe(OH)3, a reddish-brown solid.
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
Single-Displacement Reactions Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called singledisplacement reactions46, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 4.66) and the reduction of silver salts by copper (Equation 4.67 and Figure 4.21 "The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate"): Equation 4.66 Zn(s) + Fe2+(aq) → Zn2+(aq) + Fe(s) 46. A chemical reaction in which an ion in solution is displaced through oxidation of a metal.
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Equation 4.67 Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) The reaction in Equation 4.66 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object. Figure 4.21 The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate
When a copper coil is placed in a solution of silver nitrate, silver ions are reduced to metallic silver on the copper surface, and some of the copper metal dissolves. Note the formation of a metallic silver precipitate on the copper coil and a blue color in the surrounding solution due to the presence of aqueous Cu 2+ ions. (Water molecules and nitrate ions have been omitted from molecular views of the solution for clarity.)
The Activity Series By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, we saw in Equation 4.66 and Equation 4.67 that metallic zinc
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reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn 2+. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal: Equation 4.68
Zn (s) + Mg 2+ (aq) → Zn 2+ (aq) + Mg (s)
Equation 4.69 Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s) Magnesium has a greater tendency to be oxidized than zinc does. Pairwise reactions of this sort are the basis of the activity series47 (Figure 4.22 "The Activity Series"), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals48 (located at the top of the series) and which are inert metals49 (at the bottom of the series).
47. A list of metals and hydrogen in order of their relative tendency to be oxidized. 48. The metals at the top of the activity series, which have the greatest tendency to be oxidized. 49. The metals at the bottom of the activity series, which have the least tendency to be oxidized.
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Figure 4.22 The Activity Series
When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in Figure 4.22 "The Activity Series", magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example 19 demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. We will return to the activity series when we discuss oxidation–reduction reactions in more detail in Chapter 19 "Electrochemistry".
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EXAMPLE 19 Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. a. A strip of aluminum foil is placed in an aqueous solution of silver nitrate. b. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate. c. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals. Given: reactants Asked for: overall reaction and net ionic equation Strategy: A Locate the reactants in the activity series in Figure 4.22 "The Activity Series" and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced. B Write the net ionic equation for the redox reaction. Solution:
a. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows: Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s) Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction. b. A Mercury lies below lead in the activity series, so no reaction will occur.
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c. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42− form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows: Pb(s) + 2H+(aq) + SO42−(aq) → PbSO4(s) + H2(g) Lead(II) sulfate is the white solid that forms on corroded battery terminals.
Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.
Exercise Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. a. A strip of chromium metal is placed in an aqueous solution of aluminum chloride. b. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
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c. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid). Answer: a. no reaction b. 3Zn(s) + 2Cr3+(aq) → 3Zn2+(aq) + 2Cr(s) c. 2Al(s) + 6CH3CO2H(aq) → 2Al3+(aq) + 6CH3CO2−(aq) + 3H2(g)
Summary In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method"), in which the overall reaction is separated into an oxidation equation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (Figure 4.22 "The Activity Series"), which arranges metals and H2 in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series.
KEY TAKEAWAY • Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation.
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CONCEPTUAL PROBLEMS 1. Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants? 2. If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why? 3. In each redox reaction, determine which species is oxidized and which is reduced: a. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) b. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) c. BrO3−(aq) + 2MnO2(s) + H2O(l) → Br−(aq) + 2MnO4−(aq) + 2H+(aq) 4. Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is not a single-displacement reaction. 5. Of the following elements, which would you expect to have the greatest tendency to be oxidized: Zn, Li, or S? Explain your reasoning. 6. Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning. 7. Which of these metals produces H2 in acidic solution? a. b. c. d.
Ag Cd Ca Cu
8. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. a. b. c. d. e.
Mg(s) + Cu2+(aq) → Au(s) + Ag+(aq) → Cr(s) + Pb2+(aq) → K(s) + H2O(l) → Hg(l) + Pb2+(aq) →
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NUMERICAL PROBLEMS 1. Balance each redox reaction under the conditions indicated. a. b. c. d. e.
CuS(s) + NO3−(aq) → Cu2+(aq) + SO42−(aq) + NO(g); acidic solution Ag(s) + HS−(aq) + CrO42−(aq) → Ag2S(s) + Cr(OH)3(s); basic solution Zn(s) + H2O(l) → Zn2+(aq) + H2(g); acidic solution O2(g) + Sb(s) → H2O2(aq) + SbO2−(aq); basic solution UO22+(aq) + Te(s) → U4+(aq) + TeO42−(aq); acidic solution
2. Balance each redox reaction under the conditions indicated. a. b. c. d. e.
MnO4−(aq) + S2O32−(aq) → Mn2+(aq) + SO42−(aq); acidic solution Fe2+(aq) + Cr2O72−(aq) → Fe3+(aq) + Cr3+(aq); acidic solution Fe(s) + CrO42−(aq) → Fe2O3(s) + Cr2O3(s); basic solution Cl2(aq) → ClO3−(aq) + Cl−(aq); acidic solution CO32−(aq) + N2H4(aq) → CO(g) + N2(g); basic solution
3. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. a. b. c. d.
Platinum wire is dipped in hydrochloric acid. Manganese metal is added to a solution of iron(II) chloride. Tin is heated with steam. Hydrogen gas is bubbled through a solution of lead(II) nitrate.
4. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. a. b. c. d.
A few drops of NiBr2 are dropped onto a piece of iron. A strip of zinc is placed into a solution of HCl. Copper is dipped into a solution of ZnCl2. A solution of silver nitrate is dropped onto an aluminum plate.
5. Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction. 6. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu2+(aq) and nitric oxide gas.
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a. What has been oxidized? What has been reduced? b. Balance the chemical equation. 7. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: a. b. c. d.
Pt2+(aq) + Ag(s) → HCN(aq) + NaOH(aq) → Fe(NO3)3(aq) + NaOH(aq) → CH4(g) + O2(g) →
8. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: a. b. c. d.
Zn(s) + HCl(aq) → HNO3(aq) + AlCl3(aq) → K2CrO4(aq) + Ba(NO3)2(aq) → Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
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4.9 Quantitative Analysis Using Titrations LEARNING OBJECTIVE 1. To use titration methods to analyze solutions quantitatively.
50. A methodology that combines chemical reactions and stoichiometric calculations to determine the amounts or concentrations of substances present in a sample. 51. An experimental procedure in which a carefully measured volume of a solution of known concentration is added to a measured volume of a solution containing a compound whose concentration is to be determined. 52. The solution of known concentration that is reacted with a compound in a solution of unknown concentration in a titration.
To determine the amounts or concentrations of substances present in a sample, chemists use a combination of chemical reactions and stoichiometric calculations in a methodology called quantitative analysis50. Suppose, for example, we know the identity of a certain compound in a solution but not its concentration. If the compound reacts rapidly and completely with another reactant, we may be able to use the reaction to determine the concentration of the compound of interest. In a titration51, a carefully measured volume of a solution of known concentration, called the titrant52, is added to a measured volume of a solution containing a compound whose concentration is to be determined (the unknown). The reaction used in a titration can be an acid–base reaction, a precipitation reaction, or an oxidation–reduction reaction. In all cases, the reaction chosen for the analysis must be fast, complete, and specific; that is, only the compound of interest should react with the titrant. The equivalence point53 is reached when a stoichiometric amount of the titrant has been added—the amount required to react completely with the unknown.
Determining the Concentration of an Unknown Solution Using a Titration The chemical nature of the species present in the unknown dictates which type of reaction is most appropriate and also how to determine the equivalence point. The volume of titrant added, its concentration, and the coefficients from the balanced chemical equation for the reaction allow us to calculate the total number of moles of the unknown in the original solution. Because we have measured the volume of the solution that contains the unknown, we can calculate the molarity of the unknown substance. This procedure is summarized graphically here:
53. The point in a titration where a stoichiometric amount (i.e., the amount required to react completely with the unknown) of the titrant has been added.
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EXAMPLE 20 The calcium salt of oxalic acid [Ca(O2CCO2)] is found in the sap and leaves of some vegetables, including spinach and rhubarb, and in many ornamental plants. Because oxalic acid and its salts are toxic, when a food such as rhubarb is processed commercially, the leaves must be removed, and the oxalate content carefully monitored. The reaction of MnO4− with oxalic acid (HO2CCO2H) in acidic aqueous solution produces Mn2+ and CO2:
MnO 4 − (aq) + HO2 CCO2 H(aq) + H + (aq) → Mn 2+ (aq) + CO2 (g) + H2 O purple
colorless
Because this reaction is rapid and goes to completion, potassium permanganate (KMnO4) is widely used as a reactant for determining the concentration of oxalic acid. Suppose you stirred a 10.0 g sample of canned rhubarb with enough dilute H2SO4(aq) to obtain 127 mL of colorless solution. Because the added permanganate is rapidly consumed, adding small volumes of a 0.0247 M KMnO4 solution, which has a deep purple color, to the rhubarb extract does not initially change the color of the extract. When 15.4 mL of the permanganate solution have been added, however, the solution becomes a faint purple due to the presence of a slight excess of permanganate (Figure 4.23 "The Titration of Oxalic Acid with Permanganate"). If we assume that oxalic acid is the only species in solution that reacts with permanganate, what percentage of the mass of the original sample was calcium oxalate? Figure 4.23 The Titration of Oxalic Acid with Permanganate
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As small volumes of permanganate solution are added to the oxalate solution, a transient purple color appears and then disappears as the permanganate is consumed (a). As more permanganate is added, eventually all the oxalate is oxidized, and a faint purple color from the presence of excess permanganate appears, marking the endpoint (b).
Given: equation, mass of sample, volume of solution, and molarity and volume of titrant Asked for: mass percentage of unknown in sample Strategy: A Balance the chemical equation for the reaction using oxidation states. B Calculate the number of moles of permanganate consumed by multiplying the volume of the titrant by its molarity. Then calculate the number of moles of oxalate in the solution by multiplying by the ratio of the coefficients in the balanced chemical equation. Because calcium oxalate contains a 1:1 ratio of Ca2+:−O2CCO2−, the number of moles of oxalate in the solution is the same as the number of moles of calcium oxalate in the original sample. C Find the mass of calcium oxalate by multiplying the number of moles of calcium oxalate in the sample by its molar mass. Divide the mass of calcium oxalate by the mass of the sample and convert to a percentage to calculate the percentage by mass of calcium oxalate in the original sample. Solution:
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A As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using oxidation states gives 2MnO4−(aq) + 5HO2CCO2H(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Thus each mole of MnO4− added consumes 2.5 mol of oxalic acid. B Because we know the concentration of permanganate (0.0247 M) and the volume of permanganate solution that was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO 4− consumed:
15.4 mL
1 L
( 1000 mL ) (
0.0247 mol MnO 4 − 1 L
)
= 3.80 × 10 −4 mol M
The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation:
moles HO2 CCO2 H
= 3.80 × 10
−4
mol MnO 4
−
5 mol HO 2 CCO2 H 2 mol MnO 4 −
= 9.50 × 10 −4 mol HO2 CCO2 H C The problem asks for the percentage of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we need to know the mass of calcium oxalate that produced 9.50 × 10−4 mol of oxalic acid. Because calcium oxalate is Ca(O2CCO2), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction: Ca(O2CCO2)(s) + 2H+(aq) → Ca2+(aq) + HO2CCO2H(aq) The mass of calcium oxalate originally present was thus
mass of CaC2 O4
= 9.50 × 10
−4
1 mol CaC2 O4 mol HO2 CCO2 H 1 mol HO2 CCO2
= 0.122 g CaC 2 O4
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The original sample contained 0.122 g of calcium oxalate per 10.0 g of rhubarb. The percentage of calcium oxalate by mass was thus
% CaC2 O4 =
0.122 g × 100 = 1.22% 10.0 g
Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for the concentration of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this calculation. Exercise Glutathione is a low-molecular-weight compound found in living cells that is produced naturally by the liver. Health-care providers give glutathione intravenously to prevent side effects of chemotherapy and to prevent kidney problems after heart bypass surgery. Its structure is as follows:
Glutathione is found in two forms: one abbreviated as GSH (indicating the presence of an –SH group) and the other as GSSG (the disulfide form, in which
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an S–S bond links two glutathione units). The GSH form is easily oxidized to GSSG by elemental iodine: 2GSH(aq) + I2(aq) → GSSG(aq) + 2HI(aq) A small amount of soluble starch is added as an indicator. Because starch reacts with excess I2 to give an intense blue color, the appearance of a blue color indicates that the equivalence point of the reaction has been reached. Adding small volumes of a 0.0031 M aqueous solution of I 2 to 194 mL of a solution that contains glutathione and a trace of soluble starch initially causes no change. After 16.3 mL of iodine solution have been added, however, a permanent pale blue color appears because of the formation of the starch-iodine complex. What is the concentration of glutathione in the original solution? Answer: 5.2 × 10−4 M
Standard Solutions
54. A solution whose concentration is precisely known.
In Example 20, the concentration of the titrant (I2) was accurately known. The accuracy of any titration analysis depends on an accurate knowledge of the concentration of the titrant. Most titrants are first standardized; that is, their concentration is measured by titration with a standard solution54, which is a solution whose concentration is known precisely. Only pure crystalline compounds that do not react with water or carbon dioxide are suitable for use in preparing a standard solution. One such compound is potassium hydrogen phthalate (KHP), a weak monoprotic acid suitable for standardizing solutions of bases such as sodium hydroxide. The reaction of KHP with NaOH is a simple acid–base reaction. If the concentration of the KHP solution is known accurately and the titration of a NaOH solution with the KHP solution is carried out carefully, then the concentration of the NaOH solution can be calculated precisely. The standardized NaOH solution can then be used to titrate a solution of an acid whose concentration is unknown.
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Acid–Base Titrations Because most common acids and bases are not intensely colored, a small amount of an acid–base indicator is usually added to detect the equivalence point in an acid–base titration. The point in the titration at which an indicator changes color is called the endpoint55. The procedure is illustrated in Example 21.
The reaction of KHP with NaOH. As with all acid-base reactions, a salt is formed.
55. The point in a titration at which an indicator changes color.
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EXAMPLE 21 The structure of vitamin C (ascorbic acid, a monoprotic acid) is as follows:
Ascorbic acid. The upper figure shows the three-dimensional representation of ascorbic acid. Hatched lines indicate bonds that are behind the plane of the paper, and wedged lines indicate bonds that are out of the plane of the paper.
An absence of vitamin C in the diet leads to the disease known as scurvy, a breakdown of connective tissue throughout the body and of dentin in the teeth. Because fresh fruits and vegetables rich in vitamin C are readily available in developed countries today, scurvy is not a major problem. In the days of slow voyages in wooden ships, however, scurvy was common. Ferdinand Magellan, the first person to sail around the world, lost more than 90% of his crew, many to scurvy. Although a diet rich in fruits and vegetables contains more than enough vitamin C to prevent scurvy, many people take supplemental doses of vitamin C, hoping that the extra amounts will help prevent colds and other illness. Suppose a tablet advertised as containing 500 mg of vitamin C is dissolved in 100.0 mL of distilled water that contains a small amount of the acid–base indicator bromothymol blue, an indicator that is yellow in acid solution and blue in basic solution, to give a yellow solution. The addition of 53.5 mL of a 0.0520 M solution of NaOH results in a change to green at the endpoint, due
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to a mixture of the blue and yellow forms of the indicator (Figure 4.24 "The Titration of Ascorbic Acid with a Solution of NaOH"). What is the actual mass of vitamin C in the tablet? (The molar mass of ascorbic acid is 176.13 g/mol.) Figure 4.24 The Titration of Ascorbic Acid with a Solution of NaOH
The solution, containing bromothymol blue as an indicator, is initially yellow (a). The addition of a trace excess of NaOH causes the solution to turn green at the endpoint (b) and then blue.
Given: reactant, volume of sample solution, and volume and molarity of titrant Asked for: mass of unknown Strategy: A Write the balanced chemical equation for the reaction and calculate the number of moles of base needed to neutralize the ascorbic acid. B Using mole ratios, determine the amount of ascorbic acid consumed. Calculate the mass of vitamin C by multiplying the number of moles of ascorbic acid by its molar mass. Solution:
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A Because ascorbic acid acts as a monoprotic acid, we can write the balanced chemical equation for the reaction as HAsc(aq) + OH−(aq) → Asc−(aq) + H2O(l) where HAsc is ascorbic acid and Asc− is ascorbate. The number of moles of OH− ions needed to neutralize the ascorbic acid is −
moles OH = 53.5 mL
1L
( 1000 mL ) (
0.0520 mol OH − 1L
)
= 2.78 ×
B The mole ratio of the base added to the acid consumed is 1:1, so the number of moles of OH− added equals the number of moles of ascorbic acid present in the tablet:
mass ascorbic acid = 2.78 × 10 −3 mol HAsc
( 1 mol HAsc ) 176.13 g HAsc
Because 0.490 g equals 490 mg, the tablet contains about 2% less vitamin C than advertised. Exercise Vinegar is essentially a dilute solution of acetic acid in water. Vinegar is usually produced in a concentrated form and then diluted with water to give a final concentration of 4%–7% acetic acid; that is, a 4% m/v solution contains 4.00 g of acetic acid per 100 mL of solution. If a drop of bromothymol blue indicator is added to 50.0 mL of concentrated vinegar stock and 31.0 mL of 2.51 M NaOH are needed to turn the solution from yellow to green, what is the percentage of acetic acid in the vinegar stock? (Assume that the density of the vinegar solution is 1.00 g/mL.) Answer: 9.35%
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=
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Summary The concentration of a species in solution can be determined by quantitative analysis. One such method is a titration, in which a measured volume of a solution of one substance, the titrant, is added to a solution of another substance to determine its concentration. The equivalence point in a titration is the point at which exactly enough reactant has been added for the reaction to go to completion. A standard solution, a solution whose concentration is known precisely, is used to determine the concentration of the titrant. Many titrations, especially those that involve acid–base reactions, rely on an indicator. The point at which a color change is observed is the endpoint, which is close to the equivalence point if the indicator is chosen properly.
KEY TAKEAWAY • Quantitative analysis of an unknown solution can be achieved using titration methods.
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CONCEPTUAL PROBLEMS 1. The titration procedure is an application of the use of limiting reactants. Explain why this is so. 2. Explain how to determine the concentration of a substance using a titration. 3. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M acetic acid with 0.10 M NaOH, and the other corresponds to the titration of 100 mL 0.10 M NaOH with 0.10 M acetic acid. Which graph corresponds to which titration? Justify your answer.
4. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M ammonia with 0.10 M HCl, and the other corresponds to the titration of 100 mL 0.10 M NH4Cl with 0.10 M NaOH. Which graph corresponds to which titration? Justify your answer.
5. Following are two graphs that illustrate how the electrical conductivity of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M Ba(OH)2 with 0.10 M H2SO4 , and the other corresponds to the
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titration of 100 mL of 0.10 M NaOH with 0.10 M H2SO4. Which graph corresponds to which titration? Justify your answer.
ANSWERS 3. a. titration of NaOH with acetic acid b. titration of acetic acid with NaOH 5. a. titration of Ba(OH)2 with sulfuric acid b. titration of NaOH with sulfuric acid
NUMERICAL PROBLEMS 1. A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL. What is the molarity of the final solution? How many grams of KHP are in the 10.00 mL sample? 2. What volume of a 0.978 M solution of NaOH must be added to 25.0 mL of 0.583 M HCl to completely neutralize the acid? How many moles of NaOH are needed for the neutralization? 3. A student was titrating 25.00 mL of a basic solution with an HCl solution that was 0.281 M. The student ran out of the HCl solution after having added 32.46 mL, so she borrowed an HCl solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution?
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4.10 Essential Skills 3 TOPICS • Base-10 Logarithms • Calculations Using Common Logarithms
Essential Skills 1 in Chapter 1 "Introduction to Chemistry", Section 1.9 "Essential Skills 1" and Essential Skills 2 in Chapter 3 "Chemical Reactions", Section 3.7 "Essential Skills 2" described some fundamental mathematical operations used for solving problems in chemistry. This section introduces you to base-10 logarithms, a topic with which you must be familiar to do the Questions and Problems at the end of Chapter 4 "Reactions in Aqueous Solution". We will return to the subject of logarithms in Essential Skills 6 in Chapter 11 "Liquids", Section 11.9 "Essential Skills 6".
Base-10 (Common) Logarithms Essential Skills 1 introduced exponential notation, in which a base number is multiplied by itself the number of times indicated in the exponent. The number 10 3, for example, is the base 10 multiplied by itself three times (10 × 10 × 10 = 1000). Now suppose that we do not know what the exponent is—that we are given only a base of 10 and the final number. If our answer is 1000, the problem can be expressed as 10a = 1000 We can determine the value of a by using an operation called the base-10 logarithm, or common logarithm, abbreviated as log, that represents the power to which 10 is raised to give the number to the right of the equals sign. This relationship is stated as log 10a = a. In this case, the logarithm is 3 because 103 = 1000: log 103 = 3 log 1000 = 3 Now suppose you are asked to find a when the final number is 659. The problem can be solved as follows (remember that any operation applied to one side of an equality must also be applied to the other side):
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Chapter 4 Reactions in Aqueous Solution 10a = 659 log 10a = log 659 a = log 659 If you enter 659 into your calculator and press the “log” key, you get 2.819, which means that a = 2.819 and 102.819 = 659. Conversely, if you enter the value 2.819 into your calculator and press the “10x” key, you get 659. You can decide whether your answer is reasonable by comparing it with the results you get when a = 2 and a = 3: a = 2: 102 = 100 a = 2.819: 102.819 = 659 a = 3: 103 = 1000 Because the number 659 is between 100 and 1000, a must be between 2 and 3, which is indeed the case. Table 4.5 "Relationships in Base-10 Logarithms" lists some base-10 logarithms, their numerical values, and their exponential forms. Table 4.5 Relationships in Base-10 Logarithms Numerical Value Exponential Form Logarithm (a) 1000
103
3
100
102
2
10
101
1
1
100
0
0.1
10−1
−1
0.01
10−2
−2
0.001
10−3
−3
Base-10 logarithms may also be expressed as log10, in which the base is indicated as a subscript. We can write log 10a = a in either of two ways:
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Chapter 4 Reactions in Aqueous Solution log 10a = a log10 = (10a) = a The second equation explicitly indicates that we are solving for the base-10 logarithm of 10a. The number of significant figures in a logarithmic value is the same as the number of digits after the decimal point in its logarithm, so log 62.2, a number with three significant figures, is 1.794, with three significant figures after the decimal point; that is, 101.794 = 62.2, not 62.23. Skill Builder ES1 provides practice converting a value to its exponential form and then calculating its logarithm.
SKILL BUILDER ES1 Express each number as a power of 10 and then find the common logarithm. a. b. c. d. e.
10,000 0.00001 10.01 2.87 0.134
Solution a. 10,000 = 1 × 104; log 1 × 104 = 4.0 b. 0.00001 = 1 × 10−5; log 1 × 10−5 = −5.0 c. 10.01 = 1.001 × 10; log 10.01 = 1.0004 (enter 10.01 into your calculator and press the “log” key); 101.0004 = 10.01 d. 2.87 = 2.87 × 100; log 2.87 = 0.458 (enter 2.87 into your calculator and press the “log” key); 100.458 = 2.87 e. 0.134 = 1.34 × 10−1; log 0.134 = −0.873 (enter 0.134 into your calculator and press the “log” key); 10−0.873 = 0.134
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SKILL BUILDER ES2 Convert each base-10 logarithm to its numerical value. a. b. c. d. e.
3 −2.0 1.62 −0.23 −4.872
Solution a. b. c. d. e.
103 10−2 101.62 = 42 10−0.23 = 0.59 10−4.872 = 1.34 × 10−5
Calculations Using Common Logarithms Because logarithms are exponents, the properties of exponents that you learned in Essential Skills 1 apply to logarithms as well, which are summarized in Table 4.6 "Properties of Logarithms". The logarithm of (4.08 × 20.67), for example, can be computed as follows: log(4.08 × 20.67) = log 4.08 + log 20.67 = 0.611 + 1.3153 = 1.926 We can be sure that this answer is correct by checking that 10 1.926 is equal to 4.08 × 20.67, and it is. In an alternative approach, we multiply the two values before computing the logarithm: 4.08 × 20.67 = 84.3 log 84.3 = 1.926 We could also have expressed 84.3 as a power of 10 and then calculated the logarithm: log 84.3 = log(8.43 × 10) = log 8.43 + log 10 = 0.926 + 1 = 1.926
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As you can see, there may be more than one way to correctly solve a problem. We can use the properties of exponentials and logarithms to show that the logarithm of the inverse of a number (1/B) is the negative logarithm of that number (−log B):
log
1 = −log B (B )
If we use the formula for division given Table 4.6 "Properties of Logarithms" and recognize that log 1 = 0, then the logarithm of 1/B is
log
1 = log 1 − log B = 0 − log B = −log B (B )
Table 4.6 Properties of Logarithms Operation
Exponential Form
multiplication (10a)(10b) = 10a + b division
4.10 Essential Skills 3
10 a ( 10 b
Logarithm log(ab) = log a + log b
a−b log ( ba ) = log a − log b ) = 10
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SKILL BUILDER ES3 Convert each number to exponential form and then calculate the logarithm (assume all trailing zeros on whole numbers are not significant). a. b. c. d. e.
100 × 1000 0.100 ÷ 100 1000 × 0.010 200 × 3000 20.5 ÷ 0.026
Solution a. 100 × 1000 = (1 × 102)(1 × 103) log[(1 × 102)(1 × 103)] = 2.0 + 3.0 = 5.0 Alternatively, (1 × 102)(1 × 103) = 1 × 102 + 3 = 1 × 105 log(1 × 105) = 5.0 b. 0.100 ÷ 100 = (1.00 × 10−1) ÷ (1 × 102) log[(1.00 × 10−1) ÷ (1 × 102)] = 1 × 10−1−2 = 1 × 10−3 Alternatively, (1.00 × 10−1) ÷ (1 × 102) = 1 × 10[(−1) − 2] = 1 × 10−3 log(1 × 10−3) = −3.0 c. 1000 × 0.010 = (1 × 103)(1.0 × 10−2) log[(1 × 103)(1 × 10−2)] = 3.0 + (−2.0) = 1.0 Alternatively, (1 × 103)(1.0 × 10−2) = 1 × 10[3 + (−2)] = 1 × 101 log(1 × 101) = 1.0 d. 200 × 3000 = (2 × 102)(3 × 103)
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log[(2 × 102)(3 × 103)] = log(2 × 102) + log(3 × 103) = (log 2 + log 102) + (log 3 + log 103) = 0.30 + 2 + 0.48 + 3 = 5.8 Alternatively, (2 × 102)(3 × 103) = 6 × 102 + 3 = 6 × 105 log(6 × 105) = log 6 + log 105 = 0.78 + 5 = 5.8 e. 20.5 ÷ 0.026 = (2.05 × 10) ÷ (2.6 × 10−2) log[(2.05 × 10) ÷ (2.6 × 10−2)] = (log 2.05 + log 10) − (log 2.6 + log 10−2) = (0.3118 + 1) − [0.415 + (−2)] = 1.3118 + 1.585 = 2.90 Alternatively, (2.05 × 10) ÷ (2.6 × 10−2) = 0.788 × 10[1 − (−2)] = 0.788 × 103 log(0.79 × 103) = log 0.79 + log 103 = −0.102 + 3 = 2.90
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SKILL BUILDER ES4 Convert each number to exponential form and then calculate its logarithm (assume all trailing zeros on whole numbers are not significant). a. b. c. d.
10 × 100,000 1000 ÷ 0.10 25,000 × 150 658 ÷ 17
Solution a. b. c. d.
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(1 × 10)(1 × 105); logarithm = 6.0 (1 × 103) ÷ (1.0 × 10−1); logarithm = 4.00 (2.5 × 104)(1.50 × 102); logarithm = 6.57 (6.58 × 102) ÷ (1.7 × 10); logarithm = 1.59
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APPLICATION PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 3 (Section 4.1 "Aqueous Solutions"0) before proceeding to the Application Problems.Problems marked with a ♦ involve multiple concepts. 1. ♦ Acetaminophen (molar mass = 151 g/mol) is an analgesic used as a substitute for aspirin. If a child’s dose contains 80.0 mg of acetaminophen/5.00 mL of an ethanol-water solution, what is the molar concentration? Acetaminophen is frequently packaged as an ethanol-water solution rather than as an aqueous one. Why?
2. ♦ Lead may have been the first metal ever recovered from its ore by humans. Its cation (Pb2+) forms a precipitate with Cl− according to the following equation: Pb2+(aq) + 2Cl−(aq) → PbCl2(s) When PbCl2 is dissolved in hot water, its presence can be confirmed by its reaction with CrO42−, with which it forms a yellow precipitate: Pb2+(aq) + CrO42−(aq) → PbCrO4(s) The precipitate is used as a rust inhibitor and in pigments. a. What type of reaction does each equation represent? b. If 100 mL of a Pb2+ solution produce 1.65 g of lead chromate, what was the concentration of the lead solution? c. What volume of a potassium chromate solution containing 0.503 g of solute per 250.0 mL is needed for this reaction? d. If all the PbCrO4 originated from PbCl2, what volume of a 1.463 M NaCl solution was needed for the initial reaction? e. Why is there environmental concern over the use of PbCrO4? 3. ♦ Reactions that affect buried marble artifacts present a problem for archaeological chemists. Groundwater dissolves atmospheric carbon dioxide to produce an aqueous solution of carbonic acid: CO2(g) + H2O(l) → H2CO3(aq) This weakly acidic carbonic acid solution dissolves marble, converting it to soluble calcium bicarbonate:
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CaCO3(s) + H2CO3(aq) → Ca(HCO3)2(aq) Evaporation of water causes carbon dioxide to be driven off, resulting in the precipitation of calcium carbonate: Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) + CO2(g) The reprecipitated calcium carbonate forms a hard scale, or incrustation, on the surface of the object. a. If 8.5 g of calcium carbonate were obtained by evaporating 250 mL of a solution of calcium bicarbonate followed by drying, what was the molarity of the initial calcium bicarbonate solution, assuming complete reaction? b. If the overall reaction sequence was 75% efficient, how many grams of carbonic acid were initially dissolved in the 250 mL to produce the calcium bicarbonate? 4. How many Maalox tablets are needed to neutralize 5.00 mL of 0.100 M HCl stomach acid if each tablet contains 200 mg Mg(OH) 2 + 200 mg Al(OH)3? Each Rolaids tablet contains 412 mg CaCO3 + 80.0 mg Mg(OH)2. How many Rolaids tablets are needed? Suggest another formula (and approximate composition) for an effective antacid tablet. 5. Citric acid (C6H8O7, molar mass = 192.12 g/mol) is a triprotic acid extracted from citrus fruits and pineapple waste that provides tartness in beverages and jellies. How many grams of citric acid are contained in a 25.00 mL sample that requires 38.43 mL of 1.075 M NaOH for neutralization to occur? What is the formula of the calcium salt of this compound? 6. ♦ A method for determining the molarity of a strongly acidic solution has been developed based on the fact that a standard solution of potassium iodide and potassium iodate yields iodine when treated with acid: IO3− + 5I− + 6H+ → 3I2 + 3H2O Starch is used as the indicator in this titration because starch reacts with iodine in the presence of iodide to form an intense blue complex. The amount of iodine produced from this reaction can be determined by subsequent titration with thiosulfate: 2S2O32+ + I2 → S4O62− + 2I− The endpoint is reached when the solution becomes colorless. a. The thiosulfate solution was determined to be 1.023 M. If 37.63 mL of thiosulfate solution were needed to titrate a 25.00 mL sample of an acid, what was the H+ ion concentration of the acid?
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b. If the 25.00 mL sample that was titrated had been produced by dilution of a 10.00 mL sample of acid, what was the molarity of the acid in the original solution? c. Why might this be an effective method for determining the molarity of a strong acid, such as H2SO4? 7. ♦ Sewage processing occurs in three stages. Primary treatment removes suspended solids, secondary treatment involves biological processes that decompose organic matter, and tertiary treatment removes specific pollutants that arise from secondary treatment (generally phosphates). Phosphate can be removed by treating the HPO42− solution produced in the second stage with lime (CaO) to precipitate hydroxyapatite [Ca5(PO4)3OH]. a. Write a balanced chemical equation for the reaction that occurs in the tertiary treatment process. b. What has been neutralized in this process? c. Four pounds of hydroxyapatite precipitated from the water. What mass of lime was used in the reaction? d. Assuming a volume of water of 30 m3, what was the hydrogen phosphate anion concentration in the water? 8. Calcium hydroxide and calcium carbonate are effective in neutralizing the effects of acid rain on lakes. Suggest other compounds that might be effective in treating lakes. Give a plausible reason to explain why Ca(OH) 2 and CaCO3 are used. 9. ♦ Approximately 95% of the chlorine produced industrially comes from the electrolysis of sodium chloride solutions (brine): NaCl(aq) + H2O(l) → Cl2(g) + H2(g) + NaOH(aq) Chlorine is a respiratory irritant whose presence is easily detected by its penetrating odor and greenish-yellow color. One use for the chlorine produced is in the manufacture of hydrochloric acid. a. In the chemical equation shown, what has been oxidized and what has been reduced? b. Write the oxidation and reduction equations for this reaction. c. Balance the net ionic equation. d. Name another salt that might produce chlorine by electrolysis and give the expected products. Identify those products as gases, liquids, or solids. 10. ♦ The lead/acid battery used in automobiles consists of six cells that produce a 12 V electrical system. During discharge, lead(IV) oxide, lead, and aqueous sulfuric acid react to form lead(II) sulfate and water. a. What has been oxidized and what has been reduced?
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b. c. d. e.
Write and balance the equation for the reaction. What is the net ionic equation? What is the complete ionic equation? What hazard is associated with handling automobile batteries?
11. ♦ The use of iron, which is abundant in Earth’s crust, goes back to prehistoric times. In fact, it is believed that the Egyptians used iron implements approximately 5000 years ago. One method for quantifying the iron concentration in a sample involves three steps. The first step is to dissolve a portion of the sample in concentrated hydrochloric acid to produce ferric chloride; the second is to reduce Fe3+ to Fe2+ using zinc metal; and the third is to titrate Fe2+ with permanganate, producing Mn2+(aq) and ferric iron in the form of Fe2O3. a. Write chemical equations for all three steps. b. Write the net ionic equations for these three reactions. c. If 27.64 mL of a 1.032 M solution of permanganate are required to titrate 25.00 mL of Fe2+ in the third step, how many grams of Fe were in the original sample? d. Based on your answer to part c, if the original sample weighed 50.32 g, what was the percentage of iron? 12. Baking powder, which is a mixture of tartaric acid and sodium bicarbonate, is used in baking cakes and bread. Why does bread rise when you use baking powder? What type of reaction is involved?
13. An activity series exists for the halogens, which is based on the ease of reducing the diatomic halogen molecule (X2) to X−. Experimentally, it is found that fluorine is the easiest halogen to reduce (i.e., F2 is the best oxidant), and iodine is the hardest halogen to reduce (i.e., I2 is the worst oxidant). Consequently, the addition of any diatomic halogen, Y2, to solutions containing a halide ion (X−) that lies below Y in the periodic table will result in the reduction of Y 2 to Y− and the oxidation of X−. Describe what you would expect to occur when a. chlorine is added to an aqueous solution of bromide. b. iodine crystals are added to a solution of potassium bromide. Bromide is present in naturally occurring salt solutions called brines. Based on your answers, propose an effective method to remove bromide from brine.
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14. ♦ Marble is composed of mostly calcium carbonate. Assuming that acid rain contains 4.0 × 10−5 M H2SO4, approximately what volume of rain is necessary to dissolve a 250 lb marble statue? 15. ♦ One of the “first-aid” measures used to neutralize lakes whose pH has dropped to critical levels is to spray them with slaked lime (Ca(OH) 2) or limestone (CaCO3). (A slower but effective alternative is to add limestone boulders.) How much slaked lime would be needed to neutralize the acid in a lake that contains 4.0 × 10−5 M H2SO4 and has a volume of 1.2 cubic miles (5.0 × 1012 L)? 16. Recall from Section 4.3 "Stoichiometry of Reactions in Solution" that the reaction of ethanol with dichromate ion in acidic solution yields acetic acid and Cr3+(aq): Cr2O72−(aq) + CH3CH2OH(aq) → Cr3+(aq) + CH3CO2H(aq) Balance the equation for this reaction using oxidation states. (Hint: the oxidation state of carbon in the CH3 group remains unchanged, as do the oxidation states of hydrogen and oxygen.)
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ANSWERS 1. 0.106 M acetaminophen; acetaminophen is an organic compound that is much more soluble in ethanol than water, so using an ethanol/water mixture as the solvent allows a higher concentration of the drug to be used. 3. a. 0.34 M Ca(HCO3)2 b. 7.0 g H2CO3 5. 2.646 g citric acid, Ca3(C6H5O7)2 7. a. 5CaO + 3HPO42− + 2H2O → Ca5(PO4)3OH + 6OH− b. This is an acid–base reaction, in which the acid is the HPO 42− ion and the base is CaO. Transferring a proton from the acid to the base produces the PO43− ion and the hydroxide ion. c. 2.2 lbs (1 kg) of lime d. 3.6 × 10−4 M HPO42− 9. a. Chloride is oxidized, and protons are reduced. b. Oxidation: 2Cl− → Cl2 + 2e− Reduction: 2H2O + 2e− → H2 + 2OH− c. 2Cl− + 2H2O → Cl2 + H2 + 2OH−
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Chapter 5 Energy Changes in Chemical Reactions In Chapter 3 "Chemical Reactions", Section 3.3 "Chemical Equations", you learned that applying a small amount of heat to solid ammonium dichromate initiates a vigorous reaction that produces chromium(III) oxide, nitrogen gas, and water vapor. These are not the only products of this reaction that interest chemists, however; the reaction also releases energy in the form of heat and light. So our description of this reaction was incomplete. A complete description of a chemical reaction includes not only the identity, amount, and chemical form of the reactants and products but also the quantity of energy produced or consumed. In combustion reactions, heat is always a product; in other reactions, heat may be produced or consumed. This chapter introduces you to thermochemistry1, a branch of chemistry that describes the energy changes that occur during chemical reactions. In some situations, the energy produced by chemical reactions is actually of greater interest to chemists than the material products of the reaction. For example, the controlled combustion of organic molecules, primarily sugars and fats, within our cells provides the energy for physical activity, thought, and other complex chemical transformations that occur in our bodies. Similarly, our energy-intensive society extracts energy from the combustion of fossil fuels, such as coal, petroleum, and natural gas, to manufacture clothing and furniture, heat your home in winter and cool it in summer, and power the car or bus that gets you to class and to the movies. By the end of this chapter, you will know enough about thermochemistry to explain why ice cubes cool a glass of soda, how instant cold packs and hot packs work, and why swimming pools and waterbeds are heated. You will also understand what factors determine the caloric content of your diet and why even “nonpolluting” uses of fossil fuels may be affecting the environment.
1. A branch of chemistry that describes the energy changes that occur during chemical reactions.
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Thermodynamic spontaneity. The highly exothermic and dramatic thermite reaction is thermodynamically spontaneous. Reactants of aluminum and a metal oxide, usually iron, which are stable at room temperature, are ignited either in the presence of heat or by the reaction of potassium permanganate and glycerin. The resulting products are aluminum oxide, free and molten elemental metal, and a great deal of heat, which makes this an excellent method for on-site welding. Because this reaction has its own oxygen supply, it can be used for underwater welding as well.
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5.1 Energy and Work LEARNING OBJECTIVES 1. To understand the concept of energy and its various forms. 2. To know the relationship between energy, work, and heat.
2. Energy that results from atomic and molecular motion; the faster the motion, the higher the thermal energy. 3. A measure of an object’s thermal energy content. 4. One of the five forms of energy, radiant energy is carried by light, microwaves, and radio waves (the other forms of energy are thermal, chemical, nuclear, and electrical). Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. 5. One of the five forms of energy, electrical energy results from the flow of electrically charged particles. The other four forms of energy are radiant, thermal, chemical, and nuclear.
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work.
Forms of Energy The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure 5.1 "Forms of Energy"). Thermal energy2 results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature3 of an object is a measure of its thermal energy content. Radiant energy4 is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy5 results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy6 is stored in the nucleus of an atom, and chemical energy7 is stored within a chemical compound because of a particular arrangement of atoms.
6. One of the five forms of energy, nuclear energy is stored in the nucleus of an atom. The other four forms of energy are radiant, thermal, chemical, and electrical. 7. One of the five forms of energy, chemical energy is stored within a chemical compound because of a particular arrangement of atoms. The other four forms of energy are radiant, thermal, nuclear, and electrical.
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Figure 5.1 Forms of Energy
(a) Thermal energy results from atomic and molecular motion; molten steel at 2000°C has a very high thermal energy content. (b) Radiant energy (e.g., from the sun) is the energy in light, microwaves, and radio waves. (c) Lightning is an example of electrical energy, which is due to the flow of electrically charged particles. (d) Nuclear energy is released when particles in the nucleus of the atom are rearranged. (e) Chemical energy results from the particular arrangement of atoms in a chemical compound; the heat and light produced in this reaction are due to energy released during the breaking and reforming of chemical bonds.
8. Energy stored in an object because of its relative position or orientation. 9. Energy due to the motion of an
Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE)8, which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE)9 is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other.
1
object: KE = 2 mv 2 , where m is the mass of the object and v is its velocity.
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Energy can be converted from one form to another (Figure 5.2 "Interconversion of Forms of Energy") or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to mechanical work to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction. Figure 5.2 Interconversion of Forms of Energy
When a swimmer steps off the platform to dive into the water, potential energy is converted to kinetic energy. As the swimmer climbs back up to the top of the diving platform, chemical energy is converted to mechanical work.
Although energy can be converted from one form to another, the total amount of energy in the universe remains constant. This is known as the law of conservation of
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energy10.As you will learn in Chapter 18 "Chemical Thermodynamics", the law of conservation of energy is also known as the first law of thermodynamics. Energy cannot be created or destroyed.
Energy, Heat, and Work One definition of energy11 is the capacity to do work. The easiest form of work to visualize is mechanical work12 (Figure 5.3 "An Example of Mechanical Work"), which is the energy required to move an object a distance d when opposed by a force F, such as gravity: Equation 5.1
work = force × distance w = Fd Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (a), we can also write Equation 5.1 as follows:Recall from Chapter 1 "Introduction to Chemistry" that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Equation 5.2
work = mass × acceleration × distance w = mad 10. The total amount of energy in the universe remains constant. Energy can be neither created nor destroyed, but it can be converted from one form to another. 11. The capacity to do work. 12. The energy required to move an object a distance d when opposed by a force F:
w = F × d.
13. Thermal energy that can be transformed from an object at one temperature to an object at another temperature.
5.1 Energy and Work
Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the force of gravity. The amount of work done (w) and thus the energy required depends on three things: (1) the height of the second floor (the distance d); (2) your mass, which must be raised that distance against the downward acceleration due to gravity; and (3) your path, as you will learn in Section 5.2 "Enthalpy".
Figure 5.3 An Example of Mechanical Work
In contrast, heat (q)13 is thermal energy that can be transferred from an object at one temperature to an object at another temperature. The net transfer of thermal energy stops when the two objects reach the same temperature.
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The energy of an object can be changed only by the transfer of energy to or from another object in the form One form of energy is mechanical of heat,As you will learn in Chapter 6 "The Structure of work, the energy required to move an object of mass m a Atoms", hot objects can also lose energy as radiant distance d when opposed by a energy, such as heat or light. This energy is converted force F, such as gravity. to heat when it is absorbed by another object. Hence radiant energy is equivalent to heat. work performed on or by the object, or some combination of heat and work. Consider, for example, the energy stored in a fully charged battery. As shown in Figure 5.4 "Energy Transfer", this energy can be used primarily to perform work (e.g., running an electric fan) or to generate light and heat (e.g., illuminating a light bulb). When the battery is fully discharged in either case, the total change in energy is the same, even though the fraction released as work or heat varies greatly. The sum of the heat produced and the work performed equals the change in energy (ΔE): Equation 5.3
energy change = heat + work ΔE = q + w
Note the Pattern Energy can be transferred only in the form of heat, work performed on or by an object, or some combination of heat and work.
Figure 5.4 Energy Transfer
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Discharging a fully charged battery releases the same amount of energy whether the battery is used to run a fan (a) or illuminate a light bulb (b). In (a), most of the energy is used to perform work, which turns the blades of the fan and thus moves the air; only a small portion of the energy is released as heat by the electric motor. In (b), all the energy is released as heat and light; no work is done.
Energy is an extensive property of matter—for example, the amount of thermal energy in an object is proportional to both its mass and its temperature. (For more information on the properties of matter, see Chapter 1 "Introduction to Chemistry".) A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy.
Kinetic and Potential Energy The kinetic energy of an object is related to its mass m and velocity v: Equation 5.4
KE =
1 mv 2 2
For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is Equation 5.5 2 1 26.8 m 5 kg · m KE = (1360 kg) = 4.88 × 10 ( s ) 2 s2 2
14. The SI unit of energy:
1 J = 1 kg • m2 /s2 .
5.1 Energy and Work
Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J)14,The joule is named after the British physicist James Joule (1818–1889), an early worker in the field of energy. is defined as 1 kilogram·meter2/second2 (kg·m2/s2). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 103 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 105 J or 4.88 × 102 kJ. It is important to remember that the units of energy are the same regardless of the form of energy, whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same.
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To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is given by Equation 5.1 (w = Fd). According to Equation 5.2, the force (F) exerted by gravity on any object is equal to its mass (m, in this case, 1360 kg) times the acceleration (a) due to gravity (g, 9.81 m/s2 at Earth’s surface). The distance (d) is the height (h) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows: Equation 5.6
PE = Fd = mad = mgh
2 9.81 m 5 kg · m PE = (1360 kg) (36.6 m) = 4.88 × 10 ( s2 ) s2
= 4.88 × 10 5 J = 4.88 × 10 2 kJ The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero. Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage.
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A recent and spectacular example of the conversion of potential energy to kinetic energy was seen by the earthquake near the east coast of Honshu, Japan, on March 11, 2011. The magnitude 9.0 earthquake occurred along the Japan Trench subduction zone, the interface boundary between the Pacific and North American geological plates. During its westward movement, the Pacific plate became trapped under the North American plate, and its further movement was prevented. When there was sufficient potential energy to allow the Pacific plate to break free, approximately 7.1 × 1015 kJ of potential energy was released as kinetic energy, the equivalent of 4.75 × 108 tn of TNT (trinitrotoluene) or 25,003 nuclear bombs. The island of Japan experienced the worst devastation in its history from the earthquake, resulting tsunami, and aftershocks. Historical records indicate that an earthquake of such force occurs in some region of the globe approximately every 1000 years. One such earthquake and resulting tsunami is speculated to have caused the destruction of the lost city of Atlantis, referred to by the ancient Greek philosopher Plato.
Note the Pattern The units of energy are the same for all forms of energy.
Energy can also be expressed in the non-SI units of calories (cal)15, where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C.We specify the exact temperatures because the amount of energy needed to raise the temperature of 1 g of water 1°C varies slightly with elevation. To three significant figures, however, this amount is 1.00 cal over the temperature range 0°C–100°C. The name is derived from the Latin calor, meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:
15. A non-SI unit of energy: 1 cal = 4.184 J exactly.
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Equation 5.7
1 cal = 4.184 J exactly 1 J = 0.2390 cal
In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information, addressed in Section 5.4 "Thermochemistry and Nutrition".
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EXAMPLE 1 a. If the mass of a baseball is 149 g, what is the kinetic energy of a fastball clocked at 100 mi/h? b. A batter hits a pop fly, and the baseball (with a mass of 149 g) reaches an altitude of 250 ft. If we assume that the ball was 3 ft above home plate when hit by the batter, what is the increase in its potential energy? Given: mass and velocity or height Asked for: kinetic and potential energy Strategy: Use Equation 5.4 to calculate the kinetic energy and Equation 5.6 to calculate the potential energy, as appropriate. Solution:
a. The kinetic energy of an object is given by
1 2
mv 2 . In this case,
we know both the mass and the velocity, but we must convert the velocity to SI units:
v=
100 mi
( 1 h
1 h
1 min 1.61 km ) ( 60 min ) ( 60 s ) ( 1 mi ) (
The kinetic energy of the baseball is therefore
149 KE = 2
2 1 kg 44.7 m 2 kg · m g = 1.49 × 10 s2 ( 1000 g ) ( s ) 2
b. The increase in potential energy is the same as the amount of work required to raise the ball to its new altitude, which is (250 − 3) = 247 feet above its initial position. Thus
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PE = 149 g
1 kg 9.81 m 0.3048 m (247 ft ) = ( 1000 g ) ( s2 ) ( 1 ft )
Exercise a. In a bowling alley, the distance from the foul line to the head pin is 59 ft, 10 13/16 in. (18.26 m). If a 16 lb (7.3 kg) bowling ball takes 2.0 s to reach the head pin, what is its kinetic energy at impact? (Assume its speed is constant.) b. What is the potential energy of a 16 lb bowling ball held 3.0 ft above your foot? Answer: a. 3.10 × 102 J b. 65 J
KEY EQUATIONS general definition of work Equation 5.1: w = Fd Equation 5.2: w = mad relationship between energy, heat, and work Equation 5.3: ΔE = q + w kinetic energy Equation 5.4: KE
=
1 2
mv 2
potential energy in a gravitational field Equation 5.6: PE = mgh
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Summary Thermochemistry is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. Energy is the capacity to do work. Mechanical work is the amount of energy required to move an object a given distance when opposed by a force. Thermal energy is due to the random motions of atoms, molecules, or ions in a substance. The temperature of an object is a measure of the amount of thermal energy it contains. Heat (q) is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of potential energy (PE), energy caused by the relative position or orientation of an object. Kinetic energy (KE) is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the law of conservation of energy states that energy can be neither created nor destroyed. The most common units of energy are the joule (J), defined as 1 (kg·m2)/s2, and the calorie, defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J).
KEY TAKEAWAY • All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work.
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CONCEPTUAL PROBLEMS 1. What is the relationship between mechanical work and energy? 2. Does a person with a mass of 50 kg climbing a height of 15 m do work? Explain your answer. Does that same person do work while descending a mountain? 3. If a person exerts a force on an immovable object, does that person do work? Explain your answer. 4. Explain the differences between electrical energy, nuclear energy, and chemical energy. 5. The chapter describes thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy. Which form(s) of energy are represented by each of the following? a. sunlight b. the energy produced by a cathode ray tube, such as that found in a television c. the energy emitted from radioactivity d. the energy emitted from a burning candle e. the energy associated with a steam engine f. the energy emitted by a cellular phone g. the energy associated with a stick of dynamite 6. Describe the various forms of energy that are interconverted when a flashlight is switched on. 7. Describe the forms of energy that are interconverted when the space shuttle lifts off. 8. Categorize each of the following as representing kinetic energy or potential energy. a. b. c. d. e. f.
the energy associated with a laptop computer sitting on the edge of a desk shoveling snow water pouring out of a fire hydrant the energy released by an earthquake the energy in a volcano about to erupt the energy associated with a coiled spring
9. Are the units for potential energy the same as the units for kinetic energy? Can an absolute value for potential energy be obtained? Explain your answer. 10. Categorize each of the following as representing kinetic energy or potential energy.
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a. b. c. d. e.
water cascading over Niagara Falls a beaker balanced on the edge of a sink the energy released during a mudslide rollerblading the energy in a block of ice on a rooftop before a thaw
11. Why does hammering a piece of sheet metal cause the metal to heat up?
ANSWERS 3. Technically, the person is not doing any work, since the object does not move. 11. The kinetic energy of the hammer is transferred to the metal.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems. 1. Describe the mathematical relationship between (a) the thermal energy stored in an object and that object’s mass and (b) the thermal energy stored in an object and that object’s temperature. 2. How much energy (in kilojoules) is released or stored when each of the following occurs? a. A 230 lb football player is lifted to a height of 4.00 ft. b. An 11.8 lb cat jumps from a height of 6.50 ft. c. A 3.75 lb book falls off of a shelf that is 5.50 ft high. 3. Calculate how much energy (in kilojoules) is released or stored when each of the following occurs: a. A 130 lb ice skater is lifted 7.50 ft off the ice. b. A 48 lb child jumps from a height of 4.0 ft. c. An 18.5 lb light fixture falls from a 10.0 ft ceiling. 4. A car weighing 1438 kg falls off a bridge that is 211 ft high. Ignoring air resistance, how much energy is released when the car hits the water? 5. A 1 tn roller coaster filled with passengers reaches a height of 28 m before accelerating downhill. How much energy is released when the roller coaster reaches the bottom of the hill? Assume no energy is lost due to friction.
ANSWERS 1. a. The thermal energy content of an object is directly proportional to its mass. b. The thermal energy content of an object is directly proportional to its temperature. 3. a. 1.3 kJ stored b. 0.26 kJ released c. 0.251 kJ released 5. 250 kJ released
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5.2 Enthalpy LEARNING OBJECTIVES 1. To know the key features of a state function. 2. To use Hess’s law and thermochemical cycles to calculate enthalpy changes of chemical reactions.
To study the flow of energy during a chemical reaction, we need to distinguish between a system16, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings17, the rest of the universe, including the container in which the reaction is carried out (Figure 5.5 "A System and Its Surroundings"). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa. Figure 5.5 A System and Its Surroundings
16. The small, well-defined part of the universe in which we are interested. 17. All the universe that is not the system; that is, system + surroundings = universe. 18. A system that can exchange both matter and energy with its surroundings. 19. A system that can exchange energy but not matter with its surroundings.
The system is that part of the universe we are interested in studying, such as a chemical reaction inside a flask. The surroundings are the rest of the universe, including the container in which the reaction is carried out.
Three kinds of systems are important in chemistry. An open system18 can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system19 can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because
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thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system20 exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant, which must be true if energy is conserved. The state of a system21 is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function22 is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent. For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure 5.6 "Elevation as an Example of a State Function"). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose. 20. A system that can exchange neither energy nor matter with its suroundings. 21. A complete description of the system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. 22. A property of a system whose magnitude depends on only the present state of the system, not its previous history.
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Figure 5.6 Elevation as an Example of a State Function
The change in elevation between state 1 (at the bottom of the parking garage) and state 2 (at the top level of the parking garage) is the same for both paths A and B; it does not depend on which path is taken from the bottom to the top. In contrast, the distance traveled and the work needed to reach the top do depend on which path is taken. Elevation is a state function, but distance and work are not state functions.
Direction of Heat Flow The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (see chapter opening image). The balanced chemical equation for the reaction is as follows: Equation 5.8 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) We can also write this chemical equation as
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Chapter 5 Energy Changes in Chemical Reactions
Equation 5.9 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) + heat to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat (q) is transferred from a system to its surroundings is described as exothermic23. By convention, q < 0 for an exothermic reaction. When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation: Equation 5.10 heat + H2O(s) → H2O(l) When heat is transferred to a system from its surroundings, the process is endothermic24. By convention, q > 0 for an endothermic reaction.
Enthalpy of Reaction We have stated that the change in energy (ΔE) is equal to the sum of the heat produced and the work performed (Equation 5.3). Work done by an expanding gas is called pressure-volume work, also called PV work. Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows: Equation 5.11 23. A process in which heat (q) is transferred from a system to its surroundings. 24. A process in which heat (q) is transferred to a system from its surroundings.
5.2 Enthalpy
Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g) If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure 5.7 "An Example of Work Performed by a Reaction Carried Out at Constant Pressure"). The system is performing work by lifting the piston against the
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downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure) Equation 5.12 w = −PΔV The negative sign associated with PV work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy. Figure 5.7 An Example of Work Performed by a Reaction Carried Out at Constant Pressure
(a) Initially, the system (a copper penny and concentrated nitric acid) is at atmospheric pressure. (b) When the penny is added to the nitric acid, the volume of NO2 gas that is formed causes the piston to move upward to maintain the system at atmospheric pressure. In doing so, the system is performing work on its surroundings.
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The symbol E in Equation 5.3 represents the internal energy25 of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H)26 (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy E plus the product of its pressure P and volume V: Equation 5.13 H = E + PV Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. If a chemical change occurs at constant pressure (for a given P, ΔP = 0), the change in enthalpy (ΔH)27 is Equation 5.14 ΔH = Δ(E + PV) = ΔE + ΔPV = ΔE + PΔV Substituting q + w for ΔE (Equation 5.3) and −w for PΔV (Equation 5.12), we obtain Equation 5.15 25. The sum of the kinetic and potential energies of all of a system’s components. Additionally, ΔE = q + w, where q is the heat produced by the system and w is the work performed by the system. Internal energy is a state function. 26. The sum of a system’s internal energy E and the product of its pressure P and volume V :
H = E + PV.
27. At constant pressure, the amount of heat transferred from the surroundings to the system or vice versa: .ΔH = q p .
5.2 Enthalpy
ΔH = ΔE + PΔV = qp + w − w = qp The subscript p is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation 5.15 we see that at constant pressure the change in enthalpy, ΔH of the system, defined as Hfinal − Hinitial, is equal to the heat gained or lost. Equation 5.16 ΔH = Hfinal − Hinitial = qp Just as with ΔE, because enthalpy is a state function, the magnitude of ΔH depends on only the initial and final states of the system, not on the path taken. Most
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important, the enthalpy change is the same even if the process does not occur at constant pressure.
Note the Pattern To find ΔH, measure qp.
When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (ΔHrxn)28, the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so ΔHrxn is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so ΔHrxn is positive. Thus ΔHrxn < 0 for an exothermic reaction, and ΔHrxn > 0 for an endothermic reaction. In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table: Reaction Type
q
ΔHrxn
exothermic
< 0 < 0 (heat flows from a system to its surroundings)
endothermic
> 0 > 0 (heat flows from the surroundings to a system)
If ΔHrxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (part (a) in Figure 5.8 "The Enthalpy of Reaction"). Conversely, if ΔHrxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (part (b) in Figure 5.8 "The Enthalpy of Reaction"). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.
28. The change in enthalpy that occurs during a chemical reaction.
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Note the Pattern Bond breaking requires an input of energy; bond making releases energy.
• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):
Figure 5.8 The Enthalpy of Reaction
Equation 5.17
heat + H2 O (s) → H2 O (l)
ΔH > 0
Equation 5.18
H2 O (l) → H2 O (s) + heat
ΔH < 0
In both cases, the magnitude of the enthalpy change is the same; only the sign is different. Energy changes in chemical reactions are usually measured
• Enthalpy is an extensive property (like as changes in enthalpy. (a) If mass). The magnitude of ΔH for a reaction heat flows from a system to its surroundings, the enthalpy of the is proportional to the amounts of the system decreases, ΔHrxn is substances that react. For example, a large negative, and the reaction is fire produces more heat than a single exothermic; it is energetically match, even though the chemical downhill. (b) Conversely, if heat flows from the surroundings to a reaction—the combustion of wood—is the system, the enthalpy of the same in both cases. For this reason, the system increases, ΔHrxn is enthalpy change for a reaction is usually positive, and the reaction is given in kilojoules per mole of a particular endothermic; it is energetically uphill. reactant or product. Consider Equation 5.19, which describes the reaction of aluminum with iron(III) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al2O3, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe2O3 consumed:
5.2 Enthalpy
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Equation 5.19 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) + 851.5 kJ Thus ΔH = −851.5 kJ/mol of Fe2O3. We can also describe ΔH for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation 5.20, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation: Equation 5.20
2Al (s) + Fe2 O3 (s) → 2Fe (s) + Al2 O3 (s)
ΔHrxn = −851. 5 kJ
If 4 mol of Al and 2 mol of Fe2O3 react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows: Equation 5.21
−
851.5 kJ 425.8 kJ 1703 kJ =− =− 2 mol Al 1 mol Al 4 mol Al
The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example 2.
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EXAMPLE 2 Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If ΔH is 6.01 kJ/ mol for the reaction H2O(s) → H2O(l) at 0°C and constant pressure, how much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 106 metric tons)? (A metric ton is 1000 kg.) Given: energy per mole of ice and mass of iceberg Asked for: energy required to melt iceberg Strategy: A Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice. B Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice. Solution: A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by ΔH (+6.01 kJ/mol):
moles H2 O
= 1.00 × 10
6
1000 kg metric tons H2 O 1 metric ton
1000 g 1 kg
= 5.55 × 10 10 mol H2 O B The energy needed to melt the iceberg is thus
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Chapter 5 Energy Changes in Chemical Reactions
6.01 kJ mol H2 O
(5.55 × 10 10 mol H O ) = 3.34 × 10 11 kJ 2
Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below. Possible sources of the approximately 3.34 × 1011 kJ needed to melt a 1.00 × 106 metric ton iceberg Combustion of 3.8 × 103 ft3 of natural gas Combustion of 68,000 barrels of oil Combustion of 15,000 tons of coal 1.1 × 108 kilowatt-hours of electricity
Exercise If 17.3 g of powdered aluminum are allowed to react with excess Fe 2O3, how much heat is produced? Answer: 273 kJ
Hess’s Law
29. The enthalpy change (ΔH) for an overall reaction is the sum of the ΔH values for the individual reactions.
5.2 Enthalpy
Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s law29, after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.
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We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation 5.20 can be viewed as occurring in three distinct steps with known ΔH values. As shown in Figure 5.9 "Energy Changes Accompanying the Thermite Reaction", the first reaction produces 1 mol of solid aluminum oxide (Al 2O3) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation 5.22); the enthalpy change for this reaction is −732.5 kJ/mol of Fe 2O3. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 5.22); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation 5.22); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure 5.9 "Energy Changes Accompanying the Thermite Reaction", the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in Equation 5.22 gives the overall reaction, shown in part (d):
Equation 5.22
2Al(s) + Fe2 O3 (s) ⎯→
2Fe (l, 1758°C) + Al2 O3 (s)
2Fe (l, 1758°C)
⎯→
2Fe(s, 1758°C)
2Fe(s, 1758°C)
⎯→
2Fe(s, 25°C)
2Al(s) + Fe2 O3 (s) ⎯→ Al2 O3 (s) + 2Fe(s) (all at 25°C)
ΔH ΔH ΔH ΔHrxn
The net reaction in part (d) in Equation 5.22 is identical to Equation 5.20. By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure 5.9 "Energy Changes Accompanying the Thermite Reaction".
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= = = =
Chapter 5 Energy Changes in Chemical Reactions
Figure 5.9 Energy Changes Accompanying the Thermite Reaction
Because enthalpy is a state function, the overall enthalpy change for the reaction of 2 mol of Al(s) with 1 mol of Fe2O3(s) is −851.1 kJ, whether the reaction occurs in a single step (ΔH4, shown on the left) or in three hypothetical steps (shown on the right) that involve the successive formation of solid Al 2O3 and liquid iron (ΔH1), solid iron at 1758°C (ΔH2), and solid iron at 25°C (ΔH3). Thus ΔH4 = ΔH1 + ΔH2 + ΔH3, as stated by Hess’s law.
Comparing parts (a) and (d) in Equation 5.22 also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation 5.22), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation 5.22). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation. When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure:
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1. Identify the equation whose ΔH value is unknown and write individual reactions with known ΔH values that, when added together, will give the desired equation. 2. Arrange the chemical equations so that the reaction of interest is the sum of the individual reactions. 3. If a reaction must be reversed, change the sign of ΔH for that reaction. Additionally, if a reaction must be multiplied by a factor to obtain the correct number of moles of a substance, multiply its ΔH value by that same factor. 4. Add together the individual reactions and their corresponding ΔH values to obtain the reaction of interest and the unknown ΔH. We illustrate how to use this procedure in Example 3.
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EXAMPLE 3 When carbon is burned with limited amounts of oxygen gas (O 2), carbon monoxide (CO) is the main product:
(1) 2C(s) + O2 (g) → 2CO(g)
ΔH1 = −221.0 kJ
When carbon is burned in excess O2, carbon dioxide (CO2) is produced:
(2) C(s) + O2 (g) → CO2 (g)
ΔH2 = −393. 5 kJ
Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O2 to give CO2. Given: two balanced chemical equations and their ΔH values Asked for: enthalpy change for a third reaction Strategy: A After balancing the chemical equation for the overall reaction, write two equations whose ΔH values are known and that, when added together, give the equation for the overall reaction. (Reverse the direction of one or more of the equations as necessary, making sure to also reverse the sign of ΔH.) B Multiply the equations by appropriate factors to ensure that they give the desired overall chemical equation when added together. To obtain the enthalpy change per mole of CO, write the resulting equations as a sum, along with the enthalpy change for each. Solution: A We begin by writing the balanced chemical equation for the reaction of interest:
(3) CO(g) +
1 O2 (g) → CO2 (g) 2
ΔHrxn = ?
There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added
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together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give
2CO(g) → 2C(s) + O2 (g)
ΔH = +221.0 kJ
Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO 2, is the product we want in Equation 3:
C(s) + O2 (g) → CO2 (g)
ΔH2 = −393. 5 kJ
B Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant:
2C(s) + 2O2 (g) → 2CO 2 (g)
ΔH = −787.0 kJ
Writing the resulting equations as a sum, along with the enthalpy change for each, gives
2CO(g)
⎯→
2C(s) + O2 (g)
2C(s) + 2 O2 (g) ⎯→ 2CO 2 (g) 2CO(g) + O2 (g)
⎯→ 2CO 2 (g)
ΔH ΔH ΔH
= −ΔH1 = +221.0 = 2ΔH2 = −787.0 = −566.0 kJ
Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O 2, and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2:
CO(g) +
1 O2 (g) → CO2 (g) 2
ΔHrxn = −283.0 kJ/mol CO
An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms:
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Chapter 5 Energy Changes in Chemical Reactions
(A) (B) (C)
2C(s) + O2 (g)
⎯→
2CO(g)
2CO(g) + O2 (g) ⎯→ 2CO 2 (g) 2C(s) + 2O2 (g)
⎯→ 2CO 2 (g)
ΔHA ΔHB ΔHC
= ΔH1 = −221.0 kJ = ? = 2ΔH2 = 2 × ( −
The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO2. From Hess’s law, ΔHA + ΔHB = ΔHC, and we are given ΔH for reactions A and C. Substituting the appropriate values gives
−221.0 kJ + ΔHB = −787.0 kJ ΔHB = −566.0 kJ This is again the enthalpy change for the conversion of 2 mol of CO to CO 2. The enthalpy change for the conversion of 1 mol of CO to CO 2 is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem. Exercise The reaction of acetylene (C2H2) with hydrogen (H2) can produce either ethylene (C2H4) or ethane (C2H6):
C2 H2 (g) + H2 (g) → C2 H4 (g)
C2 H2 (g) + 2H2 (g) → C2 H6 (g)
ΔH = −175. 7 kJ/mol C 2 H2 ΔH = −312.0 kJ/mol C 2 H2
What is ΔH for the reaction of C2H4 with H2 to form C2H6? Answer: −136.3 kJ/mol of C2H4
Enthalpies of Formation and Reaction Chapter 2 "Molecules, Ions, and Chemical Formulas", Chapter 3 "Chemical Reactions", and Chapter 4 "Reactions in Aqueous Solution" presented a wide variety of chemical reactions, and you learned how to write balanced chemical equations that include all the reactants and the products except heat. One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the
5.2 Enthalpy
592
Chapter 5 Energy Changes in Chemical Reactions
enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following: • Enthalpy of combustion (ΔHcomb)30: Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance. • Enthalpy of fusion (ΔHfus)31: The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds. • Enthalpy of vaporization (ΔHvap)32: The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds. • Enthalpy of solution (ΔHsoln)33: The enthalpy change when a specified amount of solute dissolves in a given quantity of solvent. • Enthalpy of formation (ΔHf)34: The enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The corresponding relationship is Equation 5.23
elements → compound
ΔHrxn = ΔHf
For example, 30. The change in enthalpy that occurs during a combustion reaction. 31. The enthalpy change that acompanies the melting (fusion) of 1 mol of a substance. 32. The enthalpy change that accompanies the vaporization of 1 mol of a substance. 33. The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent.
C(s) + O2 (g) → CO 2 (g)
ΔHrxn = ΔHf [CO 2 (g)]
The sign convention for ΔHf is the same as for any enthalpy change: ΔHf < 0 if heat is released when elements combine to form a compound and ΔHf > 0 if heat is absorbed. The values of ΔHvap and ΔHfus for some common substances are listed in Table 5.1 "Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points". These values are used in enthalpy calculations when any of the substances undergoes a change of physical state during a reaction.
34. The enthalpy change for the formation of 1 mol of a compound from its component elements.
5.2 Enthalpy
593
Chapter 5 Energy Changes in Chemical Reactions
Table 5.1 Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points Substance
ΔHvap (kJ/mol) ΔHfus (kJ/mol)
argon (Ar)
6.3
1.3
methane (CH4)
9.2
0.84
ethanol (CH3CH2OH)
39.3
7.6
benzene (C6H6)
31.0
10.9
water (H2O)
40.7
6.0
mercury (Hg)
59.0
2.29
iron (Fe)
340
14
Note the Pattern The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Standard Enthalpies of Formation
35. The conditions under which most thermochemical data are tabulated: 1 atm for all gases and a concentration of 1.0 M for all species in solution. 36. The most stable form of a pure substance at a pressure of 1 atm at a specified temperature. 37. The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition.
5.2 Enthalpy
The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane. The standard conditions35 for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state36. This is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation (ΔHfο )37
594
Chapter 5 Energy Changes in Chemical Reactions
(which is pronounced “delta H eff naught”). The standard enthalpy of formation of any element in its standard state is zero by definition. For example, although oxygen can exist as ozone (O3), atomic oxygen (O), and molecular oxygen (O2), O2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H 2(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure 5.10 "Elemental Carbon"). Therefore, O2(g), H2(g), and graphite have ΔHfο values of zero. The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:
Figure 5.10 Elemental Carbon
Equation 5.24
Although graphite and diamond are both forms of elemental carbon, graphite is more stable at 1 atm pressure and 25°C than diamond is. Given enough time, diamond will revert to graphite under these conditions. Hence graphite is the standard state of carbon.
6C(s, graphite) + 6H2 (g) + 3O2 (g) → C6 H12 O6 (s)
ΔHfο = −1273.3 kJ
It is not possible to measure the value of ΔHfο for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, O2, and H2 and measuring the heat evolved as glucose is formed; the reaction shown in Equation 5.24 does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of ΔHfο are obtained using Hess’s
5.2 Enthalpy
595
Chapter 5 Energy Changes in Chemical Reactions
law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of ΔHfο for an extensive list of compounds are given in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C". Note that ΔHfο values are always reported in kilojoules per mole of the substance of interest. Also notice in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" that the standard enthalpy of formation of O2(g) is zero because it is the most stable form of oxygen in its standard state.
5.2 Enthalpy
596
Chapter 5 Energy Changes in Chemical Reactions
EXAMPLE 4 For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. a. HCl(g) b. MgCO3(s) c. CH3(CH2)14CO2H(s) (palmitic acid) Given: compound Asked for: balanced chemical equation for its formation from elements in standard states Strategy: Use Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made. Solution: To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" by a ΔHfο value of 0 kJ/mol.
a. Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are H2(g) and Cl2(g), respectively, the unbalanced chemical equation is H2(g) + Cl2(g) → HCl(g) Fractional coefficients are required in this case because ΔHfο values are reported for 1 mol of the product, HCl. Multiplying both H2(g) and Cl2(g) by 1/2 balances the equation:
5.2 Enthalpy
597
Chapter 5 Energy Changes in Chemical Reactions
1 1 H 2 (g) + Cl2 (g) → HCl(g) 2 2 b. The standard states of the elements in this compound are Mg(s), C(s, graphite), and O2(g). The unbalanced chemical equation is thus Mg(s) + C(s, graphite) + O2(g) → MgCO3(s) This equation can be balanced by inspection to give
Mg(s) + C(s, graphite) +
3 O 2 (g) → MgCO 3 (s) 2
c. Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows: C(s, graphite) + H2(g) + O2(g) → CH3(CH2)14CO2H(s) There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is 16C(s, graphite) + 16H2(g) + O2(g) → CH3(CH2)14CO2H(s)
Exercise For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. a. NaCl(s) b. H2SO4(l) c. CH3CO2H(l) (acetic acid) Answer: a. b.
5.2 Enthalpy
Na(s) + 12 Cl2 (g) → NaCl(s) H 2 (g) + 18 S 8 (s) + 2O 2 (g) → H 2 SO 4 (l)
598
Chapter 5 Energy Changes in Chemical Reactions
c. 2C(s) + O2(g) + 2H2(g) → CH3CO2H(l)
Standard Enthalpies of Reaction Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose ΔHfο values are ο 38 known. The standard enthalpy of reaction (ΔHrxn ) is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction Equation 5.25 aA + bB → cC + dD where A, B, C, and D are chemical substances and a, b, c, and d are their ο stoichiometric coefficients. The magnitude of ΔHrxn is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients: Equation 5.26 ο ΔHrxn = [cΔHfο (C) + dΔHfο (D)] − [aΔHfο (A) + bΔHfο (B)] products
reactants
More generally, we can write Equation 5.27 ο ΔHrxn = ΣmΔHfο (products) − ΣnΔHfο (reactants)
38. The enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard state.
5.2 Enthalpy
where the symbol Σ means “sum of” and m and n are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation 5.27 arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters.
599
Chapter 5 Energy Changes in Chemical Reactions
Note the Pattern “Products minus reactants” summations are typical of state functions.
To demonstrate the use of tabulated ΔHfο values, we will use them to calculate ο ΔHrxn for the combustion of glucose, the reaction that provides energy for your brain: Equation 5.28 C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Using Equation 5.27, we write Equation 5.29
ο ΔHrxn = {6ΔHfο [CO2 (g)] + 6Hfο [H2 O(l)]} − {ΔHfο [C6 H12 O6 (s)] + 6ΔH
From Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C", the relevant ΔHfο values are ΔHfο [CO2 (g)] = −393.5 kJ/mol, ΔHfο [H2 O(l)] = −285.8 kJ/mol, and ΔHfο [C6 H12 O6 (s)] = −1273.3 kJ/mol. Because O2(g) is a pure element in its standard state, ΔHfο [O2 (g)] = 0 kJ/mol. Inserting these values into Equation 5.29 and changing the subscript to indicate that this is a combustion reaction, we obtain Equation 5.30 ο ΔHcomb
= [6( − 393.5 kJ/mol) + 6( − 285.8 kJ/mol)] − [ − 1273.3 kJ/mol + 6(0 kJ/mol)] = −2802.5 kJ/mol
As illustrated in Figure 5.11 "A Thermochemical Cycle for the Combustion of ο Glucose", we can use Equation 5.29 to calculate ΔHcomb for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow ο labeled ΔHcomb . The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products
5.2 Enthalpy
600
Chapter 5 Energy Changes in Chemical Reactions
(downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the ΔHfο values of the reactants. Consequently, the enthalpy changes are
1273.3 kJ ΔH1ο = ΔHfο [glucose(s)] = −1 mol glucose 1 mol glucose 0 kJ ο ο ΔH2 = 6ΔHf [O2 (g)] = −6 mol O2 1 mol O2
= +1273
= 0 kJ
(Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change.) The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O2) to the elements is therefore +1273.3 kJ. Figure 5.11 A Thermochemical Cycle for the Combustion of Glucose
ο Two hypothetical pathways are shown from the reactants to the products. The green arrow labeled ΔHcomb indicates the combustion reaction. Alternatively, we could first convert the reactants to the elements via the reverse of the equations that define their standard enthalpies of formation (the upward arrow, labeled ΔH1ο and ΔH2ο ). Then we could convert the elements to the products via the equations used to define their standard enthalpies of ο formation (the downward arrows, labeled ΔH3ο and ΔH4ο ). Because enthalpy is a state function, ΔHcomb is ο ο ο ο equal to the sum of the enthalpy changes ΔH1 + ΔH2 + ΔH3 + ΔH4 .
The reactions that convert the elements to final products (downward purple arrows in Figure 5.11 "A Thermochemical Cycle for the Combustion of Glucose") are identical to those used to define the ΔHfο values of the products. Consequently, the
5.2 Enthalpy
601
Chapter 5 Energy Changes in Chemical Reactions
enthalpy changes (from Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C") are
−393. 5 kJ = = 6 mol CO 2 = −2361.0 kJ 1 mol CO 2 −285.8 kJ ο ο ΔH4 = 6ΔHf [H2 O(l)] = −6 mol H2 O = −1714.8 kJ 1 mol H O 2 ΔH3ο
6ΔHfο [CO2 (g)]
The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ): Equation 5.31 ο ΔHcomb = +1273.3 kJ + ( − 4075.8 kJ) = −2802.5 kJ
This is the same result we obtained using the “products minus reactants” rule (Equation 5.27) and ΔHfο values. The two results must be the same because Equation 5.29 is just a more compact way of describing the thermochemical cycle shown in Figure 5.11 "A Thermochemical Cycle for the Combustion of Glucose".
5.2 Enthalpy
602
Chapter 5 Energy Changes in Chemical Reactions
EXAMPLE 5 Long-chain fatty acids such as palmitic acid [CH3(CH2)14CO2H] are one of the two major sources of energy in our diet (ΔHfο = − 891.5 kJ/mol ). Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities ο for Chemical Substances at 25°C" to calculate ΔHcomb for the combustion of palmitic acid. Based on the energy released in combustion per gram, which is the better fuel—glucose or palmitic acid? Given: compound and ΔHfο values ο Asked for: ΔHcomb per mole and per gram
Strategy: A After writing the balanced chemical equation for the reaction, use Equation 5.27 and the values from Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate ο ΔHcomb , the energy released by the combustion of 1 mol of palmitic acid. B Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation 5.30 for the combustion of glucose to determine which is the better fuel. Solution: A To determine the energy released by the combustion of palmitic acid, we ο need to calculate its ΔHcomb . As always, the first requirement is a balanced chemical equation: C16H32O2(s) + 23O2(g) → 16CO2(g) + 16H2O(l) Using Equation 5.27 (“products minus reactants”) with ΔHfο values from Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" (and omitting the physical states of the reactants and products to save space) gives
5.2 Enthalpy
603
Chapter 5 Energy Changes in Chemical Reactions ο ΔHcomb
= ΣmΔHfο (products) − ΣnΔHfο (reactants)
= [16( − 393.5 kJ/mol CO 2 ) + 16( − 285.8 kJ/mol H 2 O] − [( − 891.5 kJ/mol C 16 H32 O2 ) + 23(0 kJ/mol O 2 )] = −9977.3 kJ/mol
This is the energy released by the combustion of 1 mol of palmitic acid. B The energy released by the combustion of 1 g of palmitic acid is ο ΔHcomb per gram =
(
9977.3 kJ 1 mol
1 mol = −38.910 kJ/g ) ( 256.42 g )
ο As calculated in Equation 5.30, ΔHcomb of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore
ο ΔHcomb per gram =
(
−2802.5 kJ 1 mol
1 mol = −15.556 kJ/g ) ( 180.16 g )
The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight. Exercise Use the data in Chapter 25 "Appendix A: Standard Thermodynamic ο Quantities for Chemical Substances at 25°C" to calculate ΔHrxn for the water–gas shift reaction, which is used industrially on an enormous scale to obtain H2(g):
CO(g) + H2 O(g) → CO2 (g) + H2 (g) water–gas shift reaction
Answer: −41.2 kJ/mol
We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s ΔHfο , which we cannot obtain otherwise. This procedure is illustrated in Example 6.
5.2 Enthalpy
604
Chapter 5 Energy Changes in Chemical Reactions
EXAMPLE 6 Beginning in 1923, tetraethyllead [(C2H5)4Pb] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are CO 2(g), H2O(l), and red PbO(s). What is the standard enthalpy of formation of ο tetraethyllead, given that ΔHcomb is −19.29 kJ/g for the combustion of ο tetraethyllead and ΔHf of red PbO(s) is −219.0 kJ/mol?
ο Given: reactant, products, and ΔHcomb values
Asked for: ΔHfο of reactant Strategy: A Write the balanced chemical equation for the combustion of tetraethyllead. Then insert the appropriate quantities into Equation 5.27 to get the equation for ΔHfο of tetraethyllead. ο ο B Convert ΔHcomb per gram given in the problem to ΔHcomb per mole by ο multiplying ΔHcomb per gram by the molar mass of tetraethyllead.
C Use Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to obtain values of ΔHfο for the other reactants and products. Insert these values into the equation for ΔHfο of tetraethyllead and solve the equation. Solution:
5.2 Enthalpy
605
Chapter 5 Energy Changes in Chemical Reactions
A The balanced chemical equation for the combustion reaction is as follows: 2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l) Using Equation 5.27 gives ο ΔHcomb
= [2ΔHfο (PbO) + 16ΔHfο (CO2 ) + 20ΔHfο (H2 O)] − {2ΔHfο [(
Solving for ΔHfο [(C2 H 5 )4 Pb] gives
ΔHfο [(C2 H5 )4 Pb] = ΔHfο (PbO) + 8ΔHfο (CO2 ) + 10ΔHfο (H2 O) − ο The values of all terms other than ΔHfο [(C2 H 5 )4 Pb] and ΔHcomb are given in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C". ο B The magnitude of ΔHcomb is given in the problem in kilojoules per gram of tetraethyllead. We must therefore multiply this value by the molar mass ο of tetraethyllead (323.44 g/mol) to get ΔHcomb for 1 mol of tetraethyllead: ο ΔHcomb =
323.44 g −19.29 kJ = −6329 kJ/mol g ( ) ( mol )
Because the balanced chemical equation contains 2 mol of tetraethyllead, ο ΔHrxn is ο ΔHrxn
−6329 kJ = 2 mol (C 2 H5 )4 Pb 1 mol (C 2 H5 )4 Pb
= −12,480 kJ
C Inserting the appropriate values into the equation for ΔHfο [(C2 H 5 )4 Pb] gives
5.2 Enthalpy
606
27 Δ 2
Chapter 5 Energy Changes in Chemical Reactions
ΔHfο [(C2 H5 )4 Pb] = [(1 mol PbO )( − 219.0 kJ/ mol PbO )] + [(8 mol CO 2 )( − 393.5 kJ/ mol CO 2 )] + [(10 mol H2 O )( − 285.8 kJ/ mol H2 O )] − −
27 mol O2 (0 kJ/ mol O2 ) [( 2 ) ] −12,480 kJ/mol (C 2 H5 )4 Pb [ ] 2
= −219.0 kJ − 3148 kJ − 2858 kJ − 0 kJ + 6240 kJ/m = 15 kJ/mol Exercise Ammonium sulfate [(NH4)2SO4] is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid: 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s) ο The value of ΔHrxn is −2805 kJ/g H2SO4. Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole).
Answer: −1181 kJ/mol
Enthalpies of Solution and Dilution Physical changes, such as melting or vaporization, and chemical reactions, in which one substance is converted to another, are accompanied by changes in enthalpy. Two other kinds of changes that are accompanied by changes in enthalpy are the dissolution of solids and the dilution of concentrated solutions. The dissolution of a solid can be described as follows:
5.2 Enthalpy
607
Chapter 5 Energy Changes in Chemical Reactions
Equation 5.32 solute(s) + solvent(l) → solution(l) The values of ΔHsoln for some common substances are given in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)". The sign and the magnitude of ΔHsoln depend on specific attractive and repulsive interactions between the solute and the solvent; these factors will be discussed in Chapter 13 "Solutions". When substances dissolve, the process can be either exothermic (ΔHsoln < 0) or endothermic (ΔHsoln > 0), as you can see from the data in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)". Table 5.2 Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol) Anion Cation
Bromide
Iodide Hydroxide
lithium
4.7
−37.0
−48.8
−63.3
−23.6
sodium
0.9
3.9
−0.6
−7.5
−44.5
potassium
−17.7
17.2
19.9
20.3
−57.6
ammonium
−1.2
14.8
16.8
13.7
—
silver
−22.5
65.5
84.4
112.2
—
magnesium
−17.7
−160.0
−185.6
−213.2
2.3
11.5
−81.3
−103.1
−119.7
−16.7
Nitrate
Acetate
lithium
−2.5
—
−18.2
−29.8
sodium
20.5
−17.3
−26.7
2.4
potassium
34.9
−15.3
−30.9
23.8
ammonium
25.7
−2.4
—
6.6
silver
22.6
—
22.6
17.8
magnesium
−90.9
—
−25.3
−91.2
calcium
−19.2
—
−13.1
−18.0
calcium
5.2 Enthalpy
Fluoride Chloride
Carbonate Sulfate
608
Chapter 5 Energy Changes in Chemical Reactions
Substances with large positive or negative enthalpies of solution have commercial applications as instant cold or hot packs. Single-use versions of these products are based on the dissolution of either calcium chloride (CaCl 2, ΔHsoln = −81.3 kJ/mol) or ammonium nitrate (NH4NO3, ΔHsoln = +25.7 kJ/mol). Both types consist of a plastic bag that contains about 100 mL of water plus a dry chemical (40 g of CaCl 2 or 30 g of NH4NO3) in a separate plastic pouch. When the pack is twisted or struck sharply, the inner plastic bag of water ruptures, and the salt dissolves in the water. If the salt is CaCl2, heat is released to produce a solution with a temperature of about 90°C; hence the product is an “instant hot compress.” If the salt is NH 4NO3, heat is absorbed when it dissolves, and the temperature drops to about 0° for an “instant cold pack.” A similar product based on the precipitation of sodium acetate, not its dissolution, is marketed as a reusable hand warmer (Figure 5.12 "An Instant Hot Pack Based on the Crystallization of Sodium Acetate"). At high temperatures, sodium acetate forms a highly concentrated aqueous solution. With cooling, an unstable supersaturated solution containing excess solute is formed. When the pack is agitated, sodium acetate trihydrate [CH3CO2Na·3H2O] crystallizes, and heat is evolved: Equation 5.33
Na+ (aq) + CH3 CO2 − (aq) + 3H2 O (l) → CH3 CO2 Na • 3H2 O (s) A bag of concentrated sodium acetate solution can be carried until heat is needed, at which time vigorous agitation induces crystallization and heat is released. The pack can be reused after it is immersed in hot water until the sodium acetate redissolves.
5.2 Enthalpy
609
Δ
Chapter 5 Energy Changes in Chemical Reactions
Figure 5.12 An Instant Hot Pack Based on the Crystallization of Sodium Acetate
The hot pack is at room temperature prior to agitation (left). Because the sodium acetate is in solution, you can see the metal disc inside the pack. After the hot pack has been agitated, the sodium acetate crystallizes (right) to release heat. Because of the mass of white sodium acetate that has crystallized, the metal disc is no longer visible.
The amount of heat released or absorbed when a substance is dissolved is not a constant; it depends on the final concentration of the solute. The ΔHsoln values given previously and in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)", for example, were obtained by measuring the enthalpy changes at various concentrations and extrapolating the data to infinite dilution. Because ΔHsoln depends on the concentration of the solute, diluting a solution can produce a change in enthalpy. If the initial dissolution process is exothermic (ΔH < 0), then the dilution process is also exothermic. This phenomenon is particularly relevant for strong acids and bases, which are often sold or stored as concentrated aqueous solutions. If water is added to a concentrated solution of sulfuric acid (which is 98% H2SO4 and 2% H2O) or sodium hydroxide, the heat released by the large negative ΔH can cause the solution to boil. Dangerous spattering of strong acid or base can be avoided if the concentrated acid or base is slowly added to water, so that the heat liberated is largely dissipated by the water. Thus you should
5.2 Enthalpy
610
Chapter 5 Energy Changes in Chemical Reactions
never add water to a strong acid or base; a useful way to avoid the danger is to remember: Add water to acid and get blasted!
KEY EQUATIONS definition of enthalpy Equation 5.11: H= E + PV pressure-volume work Equation 5.13: w = −PΔV enthalpy change at constant pressure Equation 5.14: ΔH = ΔE + PΔV Equation 5.15: ΔH = qp ο relationship between ΔHrxn and
ΔHfο
Equation 5.27:
ο ΔHrxn = ΣmΔHfο (products) − ΣnΔHfο (reactants)
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Summary In chemistry, the small part of the universe that we are studying is the system, and the rest of the universe is the surroundings. Open systems can exchange both matter and energy with their surroundings, closed systems can exchange energy but not matter with their surroundings, and isolated systems can exchange neither matter nor energy with their surroundings. A state function is a property of a system that depends on only its present state, not its history. A reaction or process in which heat is transferred from a system to its surroundings is exothermic. A reaction or process in which heat is transferred to a system from its surroundings is endothermic. Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy (ΔH) can be measured. A negative ΔH means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. For a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔHrxn. The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion (ΔHfus) and the enthalpy of vaporization (ΔHvap), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is Hess’s law. The enthalpy of combustion (ΔHcomb) is the enthalpy change that occurs when a substance is burned in excess oxygen. The enthalpy of formation (ΔHf) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation (ΔHfο ) are determined under standard conditions: a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is ο defined to be zero. The standard enthalpy of reaction (ΔHrxn ) can be calculated from the sum of the standard enthalpies of formation of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The enthalpy
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of solution (ΔHsoln) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure.
KEY TAKEAWAY • Enthalpy is a state function whose change indicates the amount of heat transferred from a system to its surroundings or vice versa, at constant pressure.
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CONCEPTUAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Conceptual Problems. 1. Heat implies the flow of energy from one object to another. Describe the energy flow in an a. exothermic reaction. b. endothermic reaction. 2. Based on the following energy diagram, a. write an equation showing how the value of ΔH2 could be determined if the values of ΔH1 and ΔH3 are known. b. identify each step as being exothermic or endothermic.
3. Based on the following energy diagram, a. write an equation showing how the value of ΔH3 could be determined if the values of ΔH1 and ΔH2 are known. b. identify each step as being exothermic or endothermic.
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4. When a thermometer is suspended in an insulated thermos that contains a block of ice, the temperature recorded on the thermometer drops. Describe the direction of heat flow. 5. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. State whether each process is endothermic or exothermic. a. Water is added to sodium hydroxide pellets, and the flask becomes hot. b. The body metabolizes glucose, producing carbon dioxide and water. c. Ammonium nitrate crystals are dissolved in water, causing the solution to become cool. 6. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. Determine whether each process is endothermic or exothermic. a. Concentrated acid is added to water in a flask, and the flask becomes warm. b. Water evaporates from your skin, causing you to shiver. c. A container of ammonium nitrate detonates. 7. Is Earth’s environment an isolated system, an open system, or a closed system? Explain your answer. 8. Why is it impossible to measure the absolute magnitude of the enthalpy of an object or a compound? 9. Determine whether energy is consumed or released in each scenario. Explain your reasoning.
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a. b. c. d. e.
A leaf falls from a tree. A motorboat maneuvers against a current. A child jumps rope. Dynamite detonates. A jogger sprints down a hill.
10. The chapter states that enthalpy is an extensive property. Why? Describe a situation that illustrates this fact. 11. The enthalpy of a system is affected by the physical states of the reactants and the products. Explain why. 12. Is the distance a person travels on a trip a state function? Why or why not? 13. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly. 14. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state? a. b. c. d.
the melting of a solid the conversion of a gas to a liquid the solidification of a liquid the dissolution of a solid into water
15. What is the difference between ΔHfο and ΔHf ? 16. In their elemental form, A2 and B2 exist as diatomic molecules. Given the following reactions, each with an associated ΔH°, describe how you would calculate ΔHfο for the compound AB2.
2AB → A 2 + B 2
3AB → AB 2 + A 2 B 2A 2 B → 2A 2 + B 2
ΔH1ο ΔH2ο ΔH3ο
17. How can ΔHfο of a compound be determined if the compound cannot be prepared by the reactions used to define its standard enthalpy of formation? 18. For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. a. HBr b. CH3OH c. NaHCO3 19. Describe the distinction between ΔHsoln and ΔHf.
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20. Does adding water to concentrated acid result in an endothermic or an exothermic process? ο 21. The following table lists ΔHsoln values for some ionic compounds. If 1 mol of each solute is dissolved in 500 mL of water, rank the resulting solutions from warmest to coldest.
Compound KOH
(kJ/mol) −57.61
LiNO3
−2.51
KMnO4
43.56
NaC2H3O2
5.2 Enthalpy
ο ΔHsoln
−17.32
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Chapter 5 Energy Changes in Chemical Reactions
NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems. 1. Using Chapter 25 "Appendix A: Standard Thermodynamic Quantities for ο Chemical Substances at 25°C", calculate ΔHrxn for each chemical reaction. a. 2Mg(s) + O2(g) → 2MgO(s) b. CaCO3(s, calcite) → CaO(s) + CO2(g) c. AgNO3(s) + NaCl(s) → AgCl(s) + NaNO3(s) 2. Using Chapter 25 "Appendix A: Standard Thermodynamic Quantities for ο Chemical Substances at 25°C", determine ΔHrxn for each chemical reaction. a. 2Na(s) + Pb(NO3)2(s) → 2NaNO3(s) + Pb(s) b. Na2CO3(s) + H2SO4(l) → Na2SO4(s) + CO2(g) + H2O(l) c. 2KClO3(s) → 2KCl(s) + 3O2(g) ο 3. Calculate ΔHrxn for each chemical equation. If necessary, balance the chemical equations.
a. Fe(s) + CuCl2(s) → FeCl2(s) + Cu(s) b. (NH4)2SO4(s) + Ca(OH)2(s) → CaSO4(s) + NH3(g) + H2O(l) c. Pb(s) + PbO2(s) + H2SO4(l) → PbSO4(s) + H2O(l) ο 4. Calculate ΔHrxn for each reaction. If necessary, balance the chemical equations.
a. 4HBr(g) + O2(g) → 2H2O(l) + 2Br2(l) b. 2KBr(s) + H2SO4(l) → K2SO4(s) + 2HBr(g) c. 4Zn(s) + 9HNO3(l) → 4Zn(NO3)2(s) + NH3(g) + 3H2O(l) 5. Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate ΔHfο of HNO3(l) if ο ΔHrxn = −320. 0 kJ for the reaction Sn(s, white) + 4HNO3(l) → SnO2(s) + 4NO2(g) + 2H2O(l). 6. Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate ΔHfο of P4O10(s) if ο ΔHrxn = −362. 1 kJ for the reaction P4O10(s) + 6H2O(l) → 4H3PO4(l). 7. How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas?
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8. How much energy is released or consumed if 10.0 g of N 2O5 is completely decomposed to produce gaseous nitrogen dioxide and oxygen? 9. In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction: NaCl(s) + H2SO4(l) + MnO2(s) → Na2SO4(s) + MnCl2(s) + H2O(l) + Cl2(g) ο Calculate ΔHrxn for this reaction. Is the reaction exothermic or endothermic?
10. Would you expect heat to be evolved during each reaction? a. solid sodium oxide with gaseous sulfur dioxide to give solid sodium sulfite b. solid aluminum chloride reacting with water to give solid aluminum oxide and hydrogen chloride gas 11. How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate? 12. Methanol is used as a fuel in Indianapolis 500 race cars. Use the following table to determine whether methanol or 2,2,4-trimethylpentane (isooctane) releases more energy per liter during combustion. ο ΔHcomb
Fuel methanol 2,2,4-trimethylpentane
(kJ/mol) Density (g/mL) −726.1
0.791
−5461.4
0.692
13. a. Use the enthalpies of combustion given in the following table to determine which organic compound releases the greatest amount of energy per gram during combustion. Fuel methanol
ο ΔHcomb
(kJ/mol) −726.1
1-ethyl-2-methylbenzene
−5210.2
n-octane
−5470.5
b. Calculate the standard enthalpy of formation of 1-ethyl-2-methylbenzene. 14. Given the enthalpies of combustion, which organic compound is the best fuel per gram?
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Fuel
ΔHfο
(kJ/mol)
ethanol
−1366.8
benzene
−3267.6
cyclooctane
−5434.7
ANSWERS 1. a. −1203 kJ/mol O2 b. 179.2 kJ c. −59.3 kJ 5. −174.1 kJ/mol 7. −20.3 kJ 9. −34.3 kJ/mol Cl2; exothermic 11. ΔH = −2.86 kJ CaCl2: −4.6 kJ; K2CO3, −0.65 kJ; mixing, −0.28 kJ 13. a. To one decimal place methanol: ΔH/g = −22.6 kJ C9H12: ΔH/g = −43.3 kJ octane: ΔH/g = −47.9 kJ Octane provides the largest amount of heat per gram upon combustion. b. ΔHf(C9H17) = −46.1 kJ/mol
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5.3 Calorimetry LEARNING OBJECTIVE 1. To use calorimetric data to calculate enthalpy changes.
Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry39 describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called calorimeters. To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object.
Heat Capacity We have seen that the temperature of an object changes when it absorbs or loses thermal energy. The magnitude of the temperature change depends on both the amount of thermal energy transferred (q) and the heat capacity of the object. Its heat capacity (C)40 is the amount of energy needed to raise the temperature of the object exactly 1°C; the units of C are joules per degree Celsius (J/°C). The change in temperature (ΔT) is Equation 5.34
ΔT =
39. A set of techniques used to measure enthalpy changes in chemical processes. 40. The amount of energy needed to raise the temperature of an object 1°C. The units of heat capacity are joules per degree Celsius (J/°C) .
q C
where q is the amount of heat (in joules), C is the heat capacity (in joules per degree Celsius), and ΔT is Tfinal − Tinitial (in degrees Celsius). Note that ΔT is always written as the final temperature minus the initial temperature. The value of C is intrinsically a positive number, but ΔT and q can be either positive or negative, and they both must have the same sign. If ΔT and q are positive, then heat flows from the surroundings into an object. If ΔT and q are negative, then heat flows from an object into its surroundings.
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The heat capacity of an object depends on both its mass and its composition. For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity (Cp)41 is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of Cp are thus J/(mol·°C).The subscript p indicates that the value was measured at constant pressure. The specific heat (Cs)42 is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g·°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change in two ways: Equation 5.35 q = nCpΔT, where n = number of moles of substance Equation 5.36 q = mCsΔT, where m = mass of substance in grams The specific heats of some common substances are given in Table 5.3 "Specific Heats of Selected Substances at 25°C". Note that the specific heat values of most solids are less than 1 J/(g·°C), whereas those of most liquids are about 2 J/(g·°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g·°C), is one of the highest known. Table 5.3 Specific Heats of Selected Substances at 25°C Compound
41. The amount of energy needed to increase the temperature of 1 mol of a substance by 1°C. The units of Cp are
J/ (mol • °C) .
42. The amount of energy needed to increase the temperature of 1 g of a substance by 1°C. The units of Cs are J/ (g • °C) .
5.3 Calorimetry
Specific Heat [J/(g·°C)]
H2O(l)
4.184
H2O(g)
2.062
CH3OH (methanol)
2.531
CH3CH2OH (ethanol)
2.438
n-C6H14 (n-hexane)
2.270
C6H6 (benzene)
1.745
C(s) (graphite)
0.709
C(s) (diamond)
0.509
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Compound
Specific Heat [J/(g·°C)]
Al(s)
0.897
Fe(s)
0.449
Cu(s)
0.385
Au(s)
0.129
Hg(l)
0.140
NaCl(s)
0.864
MgO(s)
0.921
SiO2(s) (quartz)
0.742
CaCO3(s) (calcite)
0.915
The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (Figure 5.13 "The High Specific Heat of Liquid Water Has Major Effects on Climate"). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1°C. Consequently, the mechanism for maintaining our body temperature at about 37°C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat.
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Figure 5.13 The High Specific Heat of Liquid Water Has Major Effects on Climate
Regions that are near very large bodies of water, such as oceans or lakes, tend to have smaller temperature differences between summer and winter months than regions in the center of a continent. The contours on this map show the difference between January and July monthly mean surface temperatures (in degrees Celsius).
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EXAMPLE 7 A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.)
Passive solar system. During the day (a), sunlight is absorbed by water circulating in the water wall. At night (b), heat stored in the water wall continues to warm the air inside the house.
Given: volume and density of water and initial and final temperatures Asked for: amount of energy stored Strategy: A Use the density of water at 22.0°C to obtain the mass of water (m) that corresponds to 400 L of water. Then compute ΔT for the water. B Determine the amount of heat absorbed by substituting values for m, Cs, and ΔT into Equation 5.36.
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Solution: A The mass of water is
mass of H2 O = 400 L
(
1000 mL 1 L
) ( 1 mL ) 0.998 g
= 3.99 × 10 5 g H
The temperature change (ΔT) is 38.0°C − 22.0°C = +16.0°C. B From Table 5.3 "Specific Heats of Selected Substances at 25°C", the specific heat of water is 4.184 J/(g·°C). From Equation 5.36, the heat absorbed by the water is thus
q = mCs ΔT = (3.99 × 10
5
4.184 J g ) g • °C
(16.0 °C ) = 2.67 × 10 7 J =
Both q and ΔT are positive, consistent with the fact that the water has absorbed energy. Exercise Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO2) in Table 5.3 "Specific Heats of Selected Substances at 25°C". Answer: 2.7 × 104 kJ (Even though the mass of sandstone is more than six times the mass of the water in Example 7, the amount of thermal energy stored is the same to two significant figures.)
When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process:
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Equation 5.37 qcold + qhot = 0 The equation implies that the amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite: Equation 5.38 qcold = −qhot Thus heat is conserved in any such process, consistent with the law of conservation of energy.
Note the Pattern The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object.
Substituting for q from Equation 5.36 gives Equation 5.39 [mCsΔT]hot + [mCsΔT]cold = 0 which can be rearranged to give Equation 5.40 [mCsΔT]cold = −[mCsΔT]hot When two objects initially at different temperatures are placed in contact, we can use Equation 5.40 to calculate the final temperature if we know the chemical composition and mass of the objects.
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EXAMPLE 8 If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings. Given: mass and initial temperature of two objects Asked for: final temperature Strategy: Using Equation 5.40 and writing ΔT as Tfinal − Tinitial for both the copper and the water, substitute the appropriate values of m, Cs, and Tinitial into the equation and solve for Tfinal. Solution: We can adapt Equation 5.40 to solve this problem, remembering that ΔT is defined as Tfinal − Tinitial:
[mCs (T final − T initial )]Cu + [mCs (T final − T initial )]H2 O = 0 Substituting the data provided in the problem and Table 5.3 "Specific Heats of Selected Substances at 25°C" gives
{[30.0 g][0.385 J/(g•°C)][T final − 80.0°C])} + {[100.0 g][4.184 J/(g•°C)] T final (11.6 J/°C) − 924 J + T final (418.4 J/°C) − 11,300 J T final (430 J/°C) T final
= = =
0 12,224 J 28.4°C
Exercise (a) If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings? Answer: 80.0°C Exercise (b)
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A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.) Answer: 90.6°C
Measuring Heat Flow In Example 7, radiant energy from the sun was used to raise the temperature of water. A calorimetric experiment uses essentially the same procedure, except that the thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (qrxn < 0), then heat is absorbed by the calorimeter (qcalorimeter > 0) and its temperature increases. Conversely, if the reaction absorbs heat (qrxn > 0), then heat is transferred from the calorimeter to the system (qcalorimeter < 0) and the temperature of the calorimeter decreases. In both cases, the amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction. The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants.
Constant-Pressure Calorimetry Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter43 give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure 5.14 "A Coffee-Cup Calorimeter"), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10 −6°C). Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and ΔHrxn is Equation 5.41 43. A device used to measure enthalpy changes in chemical processes at constant pressure.
5.3 Calorimetry
ΔHrxn = qrxn = −qcalorimeter = −mCsΔT The use of a constant-pressure calorimeter is illustrated in Example 9.
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Figure 5.14 A Coffee-Cup Calorimeter
This simplified version of a constant-pressure calorimeter consists of two Styrofoam cups nested and sealed with an insulated stopper to thermally isolate the system (the solution being studied) from the surroundings (the air and the laboratory bench). Two holes in the stopper allow the use of a thermometer to measure the temperature and a stirrer to mix the reactants.
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Chapter 5 Energy Changes in Chemical Reactions
EXAMPLE 9 When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is ΔHsoln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water. Given: mass of substance, volume of solvent, and initial and final temperatures Asked for: ΔHsoln Strategy: A Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution. B Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation 5.41. C Use the molar mass of KOH to calculate ΔHsoln. Solution: A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
(100.0 mL H2 O)(0.9969 g/ mL ) + 5.03 g KOH = 104.72 g The temperature change is (34.7°C − 23.0°C) = +11.7°C. B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
5.3 Calorimetry
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Chapter 5 Energy Changes in Chemical Reactions
qcalorimeter
4.184 J = mCs ΔT = (104.72 g ) g · °C
(11.7 °C ) = 5130 J =
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation 5.41, we see that ΔHrxn = −qcalorimeter = −5.13 kJ This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic. C The last step is to use the molar mass of KOH to calculate ΔHsoln—the heat released when dissolving 1 mol of KOH:
ΔHsoln =
56.11 g −5.13 kJ = −57.2 kJ/mol ( 5.03 g ) ( 1 mol )
Exercise A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example 9, find ΔHsoln for NH4Br (in kilojoules per mole). Answer: 16.6 kJ/mol
Constant-Volume Calorimetry
44. A device used to measure energy changes in chemical processes.
5.3 Calorimetry
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter44 shown schematically in Figure 5.15 "A Bomb Calorimeter". The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the
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Chapter 5 Energy Changes in Chemical Reactions
heat capacity of the bomb and the mass of water are known, the heat released can be calculated. Figure 5.15 A Bomb Calorimeter
After the temperature of the water in the insulated container has reached a constant value, the combustion reaction is initiated by passing an electric current through a wire embedded in the sample. Because this calorimeter operates at constant volume, the heat released is not precisely the same as the enthalpy change for the reaction.
Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. As you will learn in Chapter 18 "Chemical Thermodynamics", the heat released by a reaction carried out at constant volume is identical to the change in internal energy (ΔE) rather than the enthalpy change (ΔH); ΔE is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that ΔE < ΔH, the relationship between the measured temperature change and ΔHcomb is given in Equation 5.42, where Cbomb is the total heat capacity of the steel bomb and the water surrounding it:
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Chapter 5 Energy Changes in Chemical Reactions
Equation 5.42 ΔHcomb < qcomb = −qcalorimeter = −CbombΔT To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C6H5CO2H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔHcomb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation 5.42 to determine Cbomb. The use of a bomb calorimeter to measure the ΔHcomb of a substance is illustrated in Example 10.
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EXAMPLE 10 The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔHcomb of glucose?
Given: mass and ΔT for combustion of standard and sample Asked for: ΔHcomb of glucose Strategy: A Calculate the value of qrxn for benzoic acid by multiplying the mass of benzoic acid by its ΔHcomb. Then use Equation 5.42 to determine the heat capacity of the calorimeter (Cbomb) from qcomb and ΔT. B Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔHcomb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose. Solution: The first step is to use Equation 5.42 and the information obtained from the combustion of benzoic acid to calculate Cbomb. We are given ΔT, and we can calculate qcomb from the mass of benzoic acid:
qcomb = (0.579 g )( − 26.38 kJ/ g ) = −15.3 kJ From Equation 5.42,
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−Cbomb =
qcomb −15.3 kJ = = −7.34 kJ/°C ΔT 2.08°C
B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:
qcomb = −Cbomb ΔT = ( − 7.34 kJ/ °C )(3.64 °C ) = −26.7 kJ Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
ΔHcomb =
180.16 g −26.7 kJ = −2780 kJ/mol = −2.78 × 10 mol ( 1.732 g ) ( )
This result is in good agreement (< 1% error) with the value of ΔHcomb = −2803 kJ/mol that we calculated in Section 5.2 "Enthalpy" using enthalpies of formation. Exercise When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH3NHNH2) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔHcomb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.
Answer: −1.30 × 103 kJ/mol
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KEY EQUATIONS relationship of quantity of a substance, heat capacity, heat flow, and temperature change Equation 5.35: q = nCpΔT Equation 5.36: q = mCsΔT constant-pressure calorimetry Equation 5.41: ΔHrxn = qrxn = −qcalorimeter = −mCsΔT constant-volume calorimetry Equation 5.42: ΔHcomb < qcomb = −qcalorimeter = −CbombΔT
Summary Calorimetry is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called calorimeters, which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The heat capacity (C) of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The specific heat (Cs) of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the molar heat capacity (Cp) is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a constant-pressure calorimeter, which gives ΔH values directly, or a bomb calorimeter, which operates at constant volume and is particularly useful for measuring enthalpies of combustion.
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KEY TAKEAWAY • Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system.
CONCEPTUAL PROBLEMS 1. Can an object have a negative heat capacity? Why or why not? 2. What two factors determine the heat capacity of an object? Does the specific heat also depend on these two factors? Explain your answer. 3. Explain why regions along seacoasts have a more moderate climate than inland regions do. 4. Although soapstone is more expensive than brick, soapstone is frequently the building material of choice for fireplaces, particularly in northern climates with harsh winters. Propose an explanation for this.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems. 1. Using Equation 5.35 and Equation 5.36, derive a mathematical relationship between Cs and Cp. 2. Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C. Element q (J) Cp [J/(mol·K)] Final T (°C) nickel
137
silicon
26.07 19.789
3.0
zinc
603
77.5
mercury
137
57
3. Using Table 5.3 "Specific Heats of Selected Substances at 25°C", how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal. 4. Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K). a. If you put silver and gold spoons of equal mass into a cup of hot liquid and wait until the temperature of the liquid is constant, which spoon will take longer to cool down when removed from the hot liquid? b. If 8.00 g spoons of each metal at 20.0°C are placed in an insulated mug with 50.0 g of water at 97.0°C, what will be the final temperature of the water after the system has equilibrated? (Assume that no heat is transferred to the surroundings.) 5. In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy. 6. How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change?
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7. A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/ mol)" and solubility rules to identify the salt. 8. Using the data in Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds", calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)? 9. A 3.00 g sample of TNT (trinitrotoluene, C7H5N3O6) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the ΔHcomb of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the ΔHf of TNT?
ANSWERS 1. Cp = Cs × (molar mass) 3. For Cu: q = 58 J; For Al: q = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: Cs = 0.385 and 0.897 J/(g·K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al. 5. 4.7 kJ 7. ΔHsoln = −0.56 kJ/g; based on reaction with AgNO3, salt contains halide; dividing ΔHsoln values in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)" by molar mass of salts gives lithium bromide as best match, with −0.56 kJ/g. 9. Tfinal = 43.1°C; the combustion reaction is 4C7H5N3O6(s) + 21O2(g) → 28CO2(g) + 10H2O(g) + 6N2(g); ΔHfο (TNT) = −65.5 kJ/mol
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5.4 Thermochemistry and Nutrition LEARNING OBJECTIVE 1. To understand the relationship between thermochemistry and nutrition.
The thermochemical quantities that you probably encounter most often are the caloric values of food. Food supplies the raw materials that your body needs to replace cells and the energy that keeps those cells functioning. About 80% of this energy is released as heat to maintain your body temperature at a sustainable level to keep you alive. The nutritional Calorie45 (with a capital C) that you see on food labels is equal to 1 kcal (kilocalorie). The caloric content of food is determined from its enthalpy of combustion (ΔHcomb) per gram, as measured in a bomb calorimeter, using the general reaction Equation 5.43 food + excess O2(g) → CO2(g) + H2O(l) + N2(g) There are two important differences, however, between the caloric values reported for foods and the ΔHcomb of the same foods burned in a calorimeter. First, the ΔHcomb described in joules (or kilojoules) are negative for all substances that can be burned. In contrast, the caloric content of a food is always expressed as a positive number because it is stored energy. Therefore, Equation 5.44 caloric content = −ΔHcomb Second, when foods are burned in a calorimeter, any nitrogen they contain (largely from proteins, which are rich in nitrogen) is transformed to N 2. In the body, however, nitrogen from foods is converted to urea [(H2N)2C=O], rather than N2 45. A unit used to indicate the caloric content of food. It is equal to 1 kilocalorie (1 kcal).
before it is excreted. The ΔHcomb of urea measured by bomb calorimetry is −632.0 kJ/mol. Consequently, the enthalpy change measured by calorimetry for any
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nitrogen-containing food is greater than the amount of energy the body would obtain from it. The difference in the values is equal to the ΔHcomb of urea multiplied by the number of moles of urea formed when the food is broken down. This point is illustrated schematically in the following equations: Equation 5.45
food + excess O2 (g)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ CO2 (g) + ΔH1 < 0
3 O2 (g) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ 2 ΔH2 = 632.0 kJ/mol food + excess O2 (g) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
(H2 N)2 C = O(s) + excess
CO
2CO
ΔH3 = H1 + H2 <0
All three ΔH values are negative, and, by Hess’s law, ΔH3 = ΔH1 + ΔH2. The magnitude of ΔH1 must be less than ΔH3, the calorimetrically measured ΔHcomb for a food. By producing urea rather than N2, therefore, humans are excreting some of the energy that was stored in their food. Because of their different chemical compositions, foods vary widely in caloric content. As we saw in Example 5, for instance, a fatty acid such as palmitic acid produces about 39 kJ/g during combustion, while a sugar such as glucose produces 15.6 kJ/g. Fatty acids and sugars are the building blocks of fats and carbohydrates, respectively, two of the major sources of energy in the diet. Nutritionists typically assign average values of 38 kJ/g (about 9 Cal/g) and 17 kJ/g (about 4 Cal/g) for fats and carbohydrates, respectively, although the actual values for specific foods vary because of differences in composition. Proteins, the third major source of calories in the diet, vary as well. Proteins are composed of amino acids, which have the following general structure:
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In addition to their amine and carboxylic acid components, amino acids may contain a wide range of other functional groups: R can be hydrogen (–H); an alkyl group (e.g., –CH3); an aryl group (e.g., –CH2C6H5); or a substituted alkyl group that contains an amine, an alcohol, or a carboxylic acid (Figure 5.16 "The Structures of 10 Amino Acids"). Of the 20 naturally occurring amino acids, 10 are required in the human diet; these 10 are called essential amino acids because our bodies are unable to synthesize them from other compounds. Because R can be any of several different groups, each amino acid has a different value of ΔHcomb. Proteins are usually estimated to have an average ΔHcomb of 17 kJ/g (about 4 Cal/g).
General structure of an amino acid. An amino acid contains an amine group (−NH2) and a carboxylic acid group (−CO2H).
Figure 5.16 The Structures of 10 Amino Acids
Essential amino acids in this group are indicated with an asterisk.
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EXAMPLE 11 Calculate the amount of available energy obtained from the biological oxidation of 1.000 g of alanine (an amino acid). Remember that the nitrogencontaining product is urea, not N2, so biological oxidation of alanine will yield less energy than will combustion. The value of ΔHcomb for alanine is −1577 kJ/mol.
Given: amino acid and ΔHcomb per mole Asked for: caloric content per gram Strategy: A Write balanced chemical equations for the oxidation of alanine to CO 2, H2O, and urea; the combustion of urea; and the combustion of alanine. Multiply both sides of the equations by appropriate factors and then rearrange them to cancel urea from both sides when the equations are added. B Use Hess’s law to obtain an expression for ΔH for the oxidation of alanine to urea in terms of the ΔHcomb of alanine and urea. Substitute the appropriate values of ΔHcomb into the equation and solve for ΔH for the oxidation of alanine to CO2, H2O, and urea. C Calculate the amount of energy released per gram by dividing the value of ΔH by the molar mass of alanine.
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Solution: The actual energy available biologically from alanine is less than its ΔHcomb because of the production of urea rather than N2. We know the ΔHcomb values for alanine and urea, so we can use Hess’s law to calculate ΔH for the oxidation of alanine to CO2, H2O, and urea. A We begin by writing balanced chemical equations for (1) the oxidation of alanine to CO2, H2O, and urea; (2) the combustion of urea; and (3) the combustion of alanine. Because alanine contains only a single nitrogen atom, whereas urea and N2 each contain two nitrogen atoms, it is easier to balance Equations 1 and 3 if we write them for the oxidation of 2 mol of alanine:
(1) 2C3 H7 NO2 (s) + 6O2 (g) → 5CO 2 (g) + 5H2 O(l) + (H2 N)2 C=O(s) 3 (2) (H2 N)2 C=O(s) + O2 (g) → CO2 (g) + 2H2 O(l) + N2 (g) 2 15 (3) 2C 3 H7 NO2 (s) + O2 (g) → 6CO 2 (g) + 7H2 O(l) + N2 (g) 2 Adding Equations 1 and 2 and canceling urea from both sides give the overall chemical equation directly:
(1) 2C 3 H7 NO2 (s) + 6O2 (g) (2) (H2 N)2 C = O(s) + (3) 2C 3 H7 NO2 (s) +
⎯→ 5CO 2 (g) + 5H2 O(l) + (H2 N)2
3 O2 (g) ⎯→ 2
CO2 (g) + 2H2 O(l) + N
⎯→
6CO 2 (g) + 7H2 O(l) + N
15 O2 (g) 2
B By Hess’s law, ΔH3 = ΔH1 + ΔH2. We know that ΔH3 = 2ΔHcomb (alanine), ΔH2 = ΔHcomb (urea), and ΔH1 = 2ΔH (alanine → urea). Rearranging and substituting the appropriate values gives
ΔH1 = ΔH3 − ΔH2
5.4 Thermochemistry and Nutrition
= 2ΔHcomb (alanine) − ΔHcomb (urea) = 2( − 1577 kJ/mol) − ( − 632.0 kJ/mol) = −2522 kJ/(2 mol alanine)
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Thus ΔH (alanine → urea) = −2522 kJ/(2 mol of alanine) = −1261 kJ/mol of alanine. Oxidation of alanine to urea rather than to nitrogen therefore results in about a 20% decrease in the amount of energy released (−1261 kJ/ mol versus −1577 kJ/mol). C The energy released per gram by the biological oxidation of alanine is
( 1 mol
−1261 kJ
1 mol = −14.15 kJ/g ) ( 89.094 g )
This is equal to −3.382 Cal/g. Exercise Calculate the energy released per gram from the oxidation of valine (an amino acid) to CO2, H2O, and urea. Report your answer to three significant figures. The value of ΔHcomb for valine is −2922 kJ/mol.
Answer: −22.2 kJ/g (−5.31 Cal/g)
The reported caloric content of foods does not include ΔHcomb for those components that are not digested, such as fiber. Moreover, meats and fruits are 50%−70% water, which cannot be oxidized by O2 to obtain energy. So water contains no calories. Some foods contain large amounts of fiber, which is primarily composed of sugars. Although fiber can be burned in a calorimeter just like glucose to give carbon dioxide, water, and heat, humans lack the enzymes needed to break
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fiber down into smaller molecules that can be oxidized. Hence fiber also does not contribute to the caloric content of food. We can determine the caloric content of foods in two ways. The most precise method is to dry a carefully weighed sample and carry out a combustion reaction in a bomb calorimeter. The more typical approach, however, is to analyze the food for protein, carbohydrate, fat, water, and “minerals” (everything that doesn’t burn) and then calculate the caloric content using the average values for each component that produces energy (9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals). An example of this approach is shown in Table 5.4 "Approximate Composition and Fuel Value of an 8 oz Slice of Roast Beef" for a slice of roast beef. The compositions and caloric contents of some common foods are given in Table 5.5 "Approximate Compositions and Fuel Values of Some Common Foods". Table 5.4 Approximate Composition and Fuel Value of an 8 oz Slice of Roast Beef Composition
Calories
97.5 g of water
× 0 Cal/g = 0
58.7 g of protein
× 4 Cal/g = 235
69.3 g of fat
× 9 Cal/g = 624
0 g of carbohydrates × 4 Cal/g = 0 1.5 g of minerals
× 0 Cal/g = 0
Total mass: 227.0 g Total calories: about 900 Cal
Table 5.5 Approximate Compositions and Fuel Values of Some Common Foods
Food (quantity)
Approximate Composition (%)
Food Value (Cal/g)
Calories
Water Carbohydrate Protein Fat beer (12 oz)
92
3.6
0.3
0
0.4
150
coffee (6 oz)
100
~0
~0
~0
~0
~0
milk (1 cup)
88
4.5
3.3
3.3
0.6
150
egg (1 large)
75
2
12
12
1.6
80
butter (1 tbsp)
16
~0
~0
79
7.1
100
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Food (quantity)
Approximate Composition (%)
Food Value (Cal/g)
Calories
apple (8 oz)
84
15
~0
0.5
0.6
125
bread, white (2 slices)
37
48
8
4
2.6
130
brownie (40 g)
10
55
5
30
4.8
190
hamburger (4 oz)
54
0
24
21
2.9
326
fried chicken (1 drumstick)
53
8.3
22
15
2.7
195
carrots (1 cup)
87
10
1.3
~0
0.4
70
Because the Calorie represents such a large amount of energy, a few of them go a long way. An average 73 kg (160 lb) person needs about 67 Cal/h (1600 Cal/day) to fuel the basic biochemical processes that keep that person alive. This energy is required to maintain body temperature, keep the heart beating, power the muscles used for breathing, carry out chemical reactions in cells, and send the nerve impulses that control those automatic functions. Physical activity increases the amount of energy required but not by as much as many of us hope (Table 5.6 "Approximate Energy Expenditure by a 160 lb Person Engaged in Various Activities"). A moderately active individual requires about 2500−3000 Cal/day; athletes or others engaged in strenuous activity can burn 4000 Cal/day. Any excess caloric intake is stored by the body for future use, usually in the form of fat, which is the most compact way to store energy. When more energy is needed than the diet supplies, stored fuels are mobilized and oxidized. We usually exhaust the supply of stored carbohydrates before turning to fats, which accounts in part for the popularity of low-carbohydrate diets. Table 5.6 Approximate Energy Expenditure by a 160 lb Person Engaged in Various Activities Activity sleeping
Cal/h 80
driving a car
120
standing
140
eating
150
walking 2.5 mph
210
mowing lawn
250
swimming 0.25 mph
300
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Activity
Cal/h
roller skating
350
tennis
420
bicycling 13 mph
660
running 10 mph
900
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EXAMPLE 12 What is the minimum number of Calories expended by a 160 lb person who climbs a 30-story building? (Assume each flight of stairs is 14 ft high.) How many grams of glucose are required to supply this amount of energy? (The energy released during the combustion of glucose was calculated in Example 5.) Given: mass, height, and energy released by combustion of glucose Asked for: calories expended and mass of glucose needed Strategy: A Convert mass and height to SI units and then substitute these values into Equation 5.6 to calculate the change in potential energy (in kilojoules). Divide the calculated energy by 4.184 Cal/kJ to convert the potential energy change to Calories. B Use the value obtained in Example 5 for the combustion of glucose to calculate the mass of glucose needed to supply this amount of energy. Solution: The energy needed to climb the stairs equals the difference between the person’s potential energy (PE) at the top of the building and at ground level. A Recall from Section 5.1 "Energy and Work" that PE = mgh. Because m and h are given in non-SI units, we must convert them to kilograms and meters, respectively:
m = (160 lb )
14 ft h = (30 stories ) 1 story
454 g
1 kg = 72.6 kg ( 1 lb ) ( 1000 g )
12 in. 2.54 cm 1m ( 1 ft ) ( 1 in. ) ( 100 cm )
Thus
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Chapter 5 Energy Changes in Chemical Reactions PE = (72.6 kg)(9.81 m/s2)(128 m) = 8.55 × 104 (kg·m2)/s2 = 91.2 kJ To convert to Calories, we divide by 4.184 kJ/kcal:
PE = (91.2 kJ )
( 4.184 kJ ) 1 kcal
= 21.8 kcal = 21.8 Cal
B Because the combustion of glucose produces 15.6 kJ/g (Example 5), the mass of glucose needed to supply 85.5 kJ of energy is
(91.2 kJ )
( 15.6 kJ
1 g glucose
)
= 5.85 g glucose
This mass corresponds to only about a teaspoonful of sugar! Because the body is only about 30% efficient in using the energy in glucose, the actual amount of glucose required would be higher: (100%/30%) × 5.85 g = 19.5 g. Nonetheless, this calculation illustrates the difficulty many people have in trying to lose weight by exercise alone. Exercise Calculate how many times a 160 lb person would have to climb the tallest building in the United States, the 110-story Willis Tower in Chicago, to burn off 1.0 lb of stored fat. Assume that each story of the building is 14 ft high and use a calorie content of 9.0 kcal/g of fat. Answer: About 55 times
The calculations in Example 12 ignore various factors, such as how fast the person is climbing. Although the rate is irrelevant in calculating the change in potential energy, it is very relevant to the amount of energy actually required to ascend the stairs. The calculations also ignore the fact that the body’s conversion of chemical energy to mechanical work is significantly less than 100% efficient. According to the average energy expended for various activities listed in Table 5.6 "Approximate Energy Expenditure by a 160 lb Person Engaged in Various Activities", a person must run more than 4.5 h at 10 mph or bicycle for 6 h at 13 mph to burn off 1 lb of fat (1.0 lb × 454 g/lb × 9.0 Cal/g = 4100 Cal). But if a person rides a bicycle at 13 mph for only 1 h per day 6 days a week, that person will burn off 50 lb of fat in the course of a year (assuming, of course, the cyclist doesn’t increase his or her intake of calories to compensate for the exercise).
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Summary The nutritional Calorie is equivalent to 1 kcal (4.184 kJ). The caloric content of a food is its ΔHcomb per gram. The combustion of nitrogen-containing substances produces N2(g), but the biological oxidation of such substances produces urea. Hence the actual energy available from nitrogen-containing substances, such as proteins, is less than the ΔHcomb of urea multiplied by the number of moles of urea produced. The typical caloric contents for food are 9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals.
KEY TAKEAWAY • Thermochemical concepts can be applied to determine the actual energy available in food.
CONCEPTUAL PROBLEMS 1. Can water be considered a food? Explain your answer. 2. Describe how you would determine the caloric content of a bag of popcorn using a calorimeter. 3. Why do some people initially feel cold after eating a meal and then begin to feel warm? 4. In humans, one of the biochemical products of the combustion/digestion of amino acids is urea. What effect does this have on the energy available from these reactions? Speculate why conversion to urea is preferable to the generation of N2.
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NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems. 1. Determine the amount of energy available from the biological oxidation of 1.50 g of leucine (an amino acid, ΔHcomb = −3581.7 kJ/mol).
2. Calculate the energy released (in kilojoules) from the metabolism of 1.5 oz of vodka that is 62% water and 38% ethanol by volume, assuming that the total volume is equal to the sum of the volume of the two components. The density of ethanol is 0.824 g/mL. What is this enthalpy change in nutritional Calories? 3. While exercising, a person lifts an 80 lb barbell 7 ft off the ground. Assuming that the transformation of chemical energy to mechanical energy is only 35% efficient, how many Calories would the person use to accomplish this task? From Figure 5.11 "A Thermochemical Cycle for the Combustion of Glucose", how many grams of glucose would be needed to provide the energy to accomplish this task? 4. A 30 g sample of potato chips is placed in a bomb calorimeter with a heat capacity of 1.80 kJ/°C, and the bomb calorimeter is immersed in 1.5 L of water. Calculate the energy contained in the food per gram if, after combustion of the chips, the temperature of the calorimeter increases to 58.6°C from an initial temperature of 22.1°C.
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5.5 Energy Sources and the Environment LEARNING OBJECTIVE 1. To use thermochemical concepts to solve environmental issues.
Our contemporary society requires the constant expenditure of huge amounts of energy to heat our homes, provide telephone and cable service, transport us from one location to another, provide light when it is dark outside, and run the machinery that manufactures material goods. The United States alone consumes almost 106 kJ per person per day, which is about 100 times the normal required energy content of the human diet. This figure is about 30% of the world’s total energy usage, although only about 5% of the total population of the world lives in the United States. In contrast, the average energy consumption elsewhere in the world is about 105 kJ per person per day, although actual values vary widely depending on a country’s level of industrialization. In this section, we describe various sources of energy and their impact on the environment.
Fuels According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another. What is consumed on a huge scale, however, are resources that can be readily converted to a form of energy that is useful for doing work. As you will see in Chapter 18 "Chemical Thermodynamics", energy that is not used to perform work is either stored as potential energy for future use or transferred to the surroundings as heat. A major reason for the huge consumption of energy by our society is the low efficiency of most machines in transforming stored energy into work. Efficiency can be defined as the ratio of useful work accomplished to energy expended. Automobiles, for example, are only about 20% efficient in converting the energy stored in gasoline to mechanical work; the rest of the energy is released as heat, either emitted in the exhaust or produced by friction in bearings and tires. The production of electricity by coal- or oil-powered steam turbines is significantly more efficient (Figure 5.17 "Electricity from Coal"): about 38% of the energy released from combustion is converted to electricity. In comparison, modern nuclear power plants can be more than 50% efficient.
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Figure 5.17 Electricity from Coal
A coal-powered electric power plant uses the combustion of coal to produce steam, which drives a turbine to produce electricity.
In general, it is more efficient to use primary sources of energy directly (such as natural gas or oil) than to transform them to a secondary source such as electricity prior to their use. For example, if a furnace is well maintained, heating a house with natural gas is about 70% efficient. In contrast, burning the natural gas in a remote power plant, converting it to electricity, transmitting it long distances through wires, and heating the house by electric baseboard heaters have an overall efficiency of less than 35%. The total expenditure of energy in the world each year is about 3 × 1017 kJ. More than 80% of this energy is provided by the combustion of fossil fuels: oil, coal, and natural gas. (The sources of the energy consumed in the United States in 2009 are shown in Figure 5.18 "Energy Consumption in the United States by Source, 2009".) Natural gas and petroleum, whose compositions were described in Chapter 2 "Molecules, Ions, and Chemical Formulas", are the preferred fuels because they or products derived from them are gases or liquids that are readily transported, stored, and burned. Natural gas and petroleum are derived from the remains of marine creatures that died hundreds of millions of years ago and were buried beneath layers of sediment. As the sediment turned to rock, the tremendous heat and pressure inside Earth transformed the organic components of the buried sea creatures to petroleum and natural gas.
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Coal Coal46 is a complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. Because plants contain large amounts of cellulose, derived from linked glucose units, the structure of coal is more complex than that of petroleum (Figure 5.19 "The Structures of Cellulose and Coal"). In particular, coal contains a large number of oxygen atoms that link parts of the structure together, in addition to the basic framework of carbon–carbon bonds. It is impossible to draw a single structure for coal; however, because of the prevalence of rings of carbon atoms (due to the original high cellulose content), coal is more similar to an aromatic hydrocarbon than an aliphatic one.
Figure 5.18 Energy Consumption in the United States by Source, 2009
More than 80% of the total energy expended is provided by the combustion of fossil fuels, such as oil, coal, and natural gas.
Figure 5.19 The Structures of Cellulose and Coal
(a) Cellulose consists of long chains of cyclic glucose molecules linked by hydrogen bonds. (b) When cellulose is subjected to high pressures and temperatures for long periods of time, water is eliminated, and bonds are formed between the rings, eventually producing coal. This drawing shows some of the common structural features of coal; note the presence of many different kinds of ring structures.
46. A complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. It is used as a fuel.
There are four distinct classes of coal (Table 5.7 "Properties of Different Types of Coal"); their hydrogen and oxygen contents depend on the length of time the coal has been buried and the pressures and temperatures to which it has been subjected. Lignite, with a hydrogen:carbon ratio of about 1.0 and a high oxygen content, has the lowest ΔHcomb. Anthracite, in contrast, with a hydrogen:carbon ratio of about 0.5 and the lowest oxygen content, has the highest ΔHcomb and is the highest grade
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of coal. The most abundant form in the United States is bituminous coal, which has a high sulfur content because of the presence of small particles of pyrite (FeS 2). As discussed in Chapter 4 "Reactions in Aqueous Solution", the combustion of coal releases the sulfur in FeS2 as SO2, which is a major contributor to acid rain. Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds" compares the ΔHcomb per gram of oil, natural gas, and coal with those of selected organic compounds. Table 5.7 Properties of Different Types of Coal
Type
% Hydrogen:Carbon % % Heat Carbon Mole Ratio Oxygen Sulfur Content
US Deposits
anthracite
92
0.5
3
1 high
Pennsylvania, New York
bituminous
80
0.6
8
5 medium
Appalachia, Midwest, Utah
subbituminous
77
0.9
16
1 medium
Rocky Mountains
lignite
71
1.0
23
1 low
Montana
Table 5.8 Enthalpies of Combustion of Common Fuels and Selected Organic Compounds Fuel dry wood
ΔHcomb (kJ/g) −15
peat
−20.8
bituminous coal
−28.3
charcoal
−35
kerosene
−37
C6H6 (benzene)
−41.8
crude oil
−43
natural gas
−50
C2H2 (acetylene)
−50.0
CH4 (methane)
−55.5
gasoline
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Fuel hydrogen
ΔHcomb (kJ/g) −143
Peat, a precursor to coal, is the partially decayed remains of plants that grow in swampy areas. It is removed from the ground in the form of soggy bricks of mud that will not burn until they have been dried. Even though peat is a smoky, poorburning fuel that gives off relatively little heat, humans have burned it since ancient times (Figure 5.20 "A Peat Bog"). If a peat bog were buried under many layers of sediment for a few million years, the peat could eventually be compressed and heated enough to become lignite, the lowest grade of coal; given enough time and heat, lignite would eventually become anthracite, a much better fuel.
Converting Coal to Gaseous and Liquid Fuels Figure 5.20 A Peat Bog
Oil and natural gas resources are limited. Current estimates suggest that the known reserves of petroleum will be exhausted in about 60 years, and supplies of natural gas are estimated to run out in about 120 years. Coal, on the other hand, is relatively abundant, making up more than 90% of the world’s fossil fuel reserves. As a solid, coal is much more difficult to mine and ship than petroleum (a liquid) or natural gas. Consequently, Peat is a smoky fuel that burns more than 75% of the coal produced each year is simply poorly and produces little heat, but it has been used as a fuel burned in power plants to produce electricity. A great deal of current research focuses on developing methods since ancient times. to convert coal to gaseous fuels (coal gasification) or liquid fuels (coal liquefaction). In the most common approach to coal gasification, coal reacts with steam to produce a mixture of CO and H2 known as synthesis gas, or syngas:Because coal is 70%–90% carbon by mass, it is approximated as C in Equation 5.46. Equation 5.46
C(s) + H2 O(g) → CO(g) + H2 (g)
ΔH = 131 kJ
Converting coal to syngas removes any sulfur present and produces a clean-burning mixture of gases. Syngas is also used as a reactant to produce methane and methanol. A promising approach is to convert coal directly to methane through a series of reactions:
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Equation 5.47
Overall:
2C(s) + 2H2 O(g) → 2CO(g) + 2H2 (g)
ΔH1
= 262 k
CO(g) + H2 O(g) → CO2 (g) + H2 (g)
ΔH2
= −41 k
CO(g) + 3H2 (g) → CH4 (g) + H2 O(g)
ΔH3
= −206
2C(s) + 2H2 O(g) → CH4 (g) + CO2 (g)
ΔHcomb
Burning a small amount of coal or methane provides the energy consumed by these reactions. Unfortunately, methane produced by this process is currently significantly more expensive than natural gas. As supplies of natural gas become depleted, however, this coal-based process may well become competitive in cost. Similarly, the techniques available for converting coal to liquid fuels are not yet economically competitive with the production of liquid fuels from petroleum. Current approaches to coal liquefaction use a catalyst to break the complex network structure of coal into more manageable fragments. The products are then treated with hydrogen (from syngas or other sources) under high pressure to produce a liquid more like petroleum. Measuring crude oil. The Subsequent distillation, cracking, and reforming can be standard industrial unit of measure for crude oil is the 42 gal used to create products similar to those obtained from barrel. petroleum. (For more information about cracking, see Chapter 2 "Molecules, Ions, and Chemical Formulas", Section 2.6 "Industrially Important Chemicals".) The total yield of liquid fuels is about 5.5 bbl of crude liquid per ton of coal (1 bbl is 42 gal or 160 L). Although the economics of coal liquefaction are currently even less attractive than for coal gasification, liquid fuels based on coal are likely to become economically competitive as supplies of petroleum are consumed.
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= 15
Chapter 5 Energy Changes in Chemical Reactions
